From c630304bdda64498d4d2eb3ea14143ba7f03bf1b Mon Sep 17 00:00:00 2001 From: Eric Swenson Date: Thu, 4 Sep 2025 09:26:34 -0700 Subject: [PATCH] Adding two versions of HAKMEM. Setting date for HAKMEM 141 to be 1972-03-16. Setting date for HAKMEM 17 to be 1971-01-01, although this date is probably not correct, but we have no info to base its date on other than that it was "before" 1972. --- Makefile | 3 +- build/timestamps.txt | 2 + doc/mb/hakmem.141 | 3513 ++++++++++++++++++++++++++++++++++++++++++ doc/mb/hakmem.17 | 1029 +++++++++++++ 4 files changed, 4546 insertions(+), 1 deletion(-) create mode 100644 doc/mb/hakmem.141 create mode 100644 doc/mb/hakmem.17 diff --git a/Makefile b/Makefile index 344865d7..ac99f395 100644 --- a/Makefile +++ b/Makefile @@ -63,7 +63,8 @@ DOC = info _info_ sysdoc sysnet syshst kshack _teco_ emacs emacs1 c kcc \ aplogo _temp_ pdp11 chsncp cbf rug bawden llogo eak clib teach pcnet \ combat pdl minits mits_s chaos hal -pics- imlac maint cent ksc klh \ digest prs decus bsg madman hur lmdoc rrs danny netwrk klotz hello \ - clu r mini nova sits jay rjl nlogo mprog2 mudbug cfs hudini shrdlu + clu r mini nova sits jay rjl nlogo mprog2 mudbug cfs hudini shrdlu \ + mb BIN = sys sys1 sys2 emacs _teco_ lisp liblsp alan sail comlap \ c decsys graphs draw datdrw fonts fonts1 fonts2 games macsym \ maint _www_ gt40 llogo bawden sysbin -pics- lmman shrdlu imlac \ diff --git a/build/timestamps.txt b/build/timestamps.txt index 316565db..0b2f24e2 100644 --- a/build/timestamps.txt +++ b/build/timestamps.txt @@ -1795,6 +1795,8 @@ maxtul/strmrg.70 198107181941.49 maxtul/timepn.1 198012121633.43 maxtul/toolm.1 198107151807.50 mb/titler.3 197109212338.08 +mb/hakmem.141 197203160000.00 +mb/hakmem.17 197101010000.00 mbprog/assem.fbin 197902031659.44 mbprog/assem.nbin 197607140800.08 mbprog/bigpri.nbin 197906211101.06 diff --git a/doc/mb/hakmem.141 b/doc/mb/hakmem.141 new file mode 100644 index 00000000..5d380808 --- /dev/null +++ b/doc/mb/hakmem.141 @@ -0,0 +1,3513 @@ +Compiled with the hope that a record of the +random things people do around here can save +some duplication of effort -- except for fun. + +Here is some little known data which may be of interest to +computer hackers. The items and examples are so sketchy that to +decipher them may require more sincerity and curiosity than a +non-hacker can muster. Doubtless, little of this is new, but +nowadays it's hard to tell. So we must be content to give you an +insight, or save you some cycles, and to welcome further +contributions of items, new or used. + +The classification of items into sections is even more illogical +than necessary. This is because later elaborations tend to shift +perspective on many items, and this elaboration will (hopefully) +continue after publication, since this text is retained in +"machinable" form. We forgive in advance anyone deterred by this +wretched typography. + +People referred to are +from the A. I. Lab: + Marvin Minsky Rich Schroeppel + Bill Gosper Michael Speciner + Michael Beeler Gerald Sussman + John Roe Joe Cohen + Richard Stallman David Waltz + Jerry Freiberg David Silver + +once at the A. I. Lab but now elsewhere: + Jan Kok William Henneman + Rici Liknaitzky George Mitchell + Peter Samson Stuart Nelson + Roger Banks Rollo Silver + Mike Paterson + +at Digital Equipment Corporation: + Jud Lenard Dave Plumer + Ben Gurley (deceased) Steve Root + +elsewhere at M.I.T.: + Gene Salamin PDP-1 hackers + Eric Jensen Frances Yao + Edward Fredkin + +once at M.I.T. but now elsewhere: + Jackson Wright Steve Brown + Malcolm Rayfield + +at Computer Corporation of America: + Bill Mann + +in France: + Marco Schutzenberger Henry Cohen + +at BBN: + Robert Clements + CAVEATS: +Some of this material is very inside -- many readers will have to +excuse cryptic references. + +The label "PROBLEM" does not always mean exercise; +if no solution is given, it means we couldn't solve it. +If you solve a problem in here, let us know. + +Unless otherwise stated, all computer programs are in PDP-6/10 +assembly language. + +CONTENTS, HAKMEM 140 + + 4 GEOMETRY, ALGEBRA, CALCULUS + 7 RECURRENCE RELATIONS +11 BOOLEAN ALGEBRA +13 RANDOM NUMBERS +14 NUMBER THEORY, PRIMES, PROBABILITY +23 AUTOMATA THEORY +24 GAMES +26 PROPOSED COMPUTER PROGRAMS +29 CONTINUED FRACTIONS +30 GROUP THEORY +30 SET THEORY +31 QUATERNIONS +33 POLYOMINOS, ETC. +36 TOPOLOGY +39 SERIES +43 FLOWS AND ITERATED FUNCTIONS +45 PI +50 PROGRAMMING HACKS +60 PROGRAMMING ALGORITHMS, HEURISTICS +64 HARDWARE + note: this page will not be included in final publication; +it is for inter-hacker notes, etc. + +INCLUDE: C CURVE gscale-invrse mirr, PEGSOLITAIRE self dual, XE+YPI, PUSH Q,6(R), +IO PHILOS., SPECINER NU FCN SUMS, SLIVER BOX ILLUSION uses, Y = A -1/X, +(X+Y)/(1+XY), HEWITT'S CLAIM-- +R'S ALG FOR REGEXPR SERCH, POSSIBLE WINDOWING IN RECURRENCES, +Hilbert's tenth with reference, Recent history of 4 color theorem, +Risch on integration, Baker on effective solution of Diophantine eqs. +1 26 66 26 1 genfn, PERSPEC PROB., cf gamma, +cf hype (recip vs 10* etc),1/pi vs pi, bignum rep--quoti estm, +Approximants to 2^(n/12), Primes < 2^n, COROUTINES, CONTINUED FRACTION +ARITHMETIC ..6 0 9! gen cf techn--strang, cf interval trick, +LINEAR MEDIAN COMP?, g's sin & cos routines, +g: is bunchy sort (xor to get rel bit) in? -- mb, commut polyns frm higher +recurrence ntupication?, bum hamming aos, 36 cnt, fix bridgehands, 2nots +ES recrsv FFT, div by 0 trick, cf pi+1/pi, vis noise, jensen -digits, +Minsky's Venn diagram, rewrite para in search program +254 * 3937 (r) + ********************************************* +GEOMETRY, ALGEBRA, CALCULUS +********************************************* + +ITEM (Schroeppel): +(1/3)! and (2/3)! are interexpressible. +(1/4)! and (3/4)! are interexpressible. +Thus these two pairs are of dimensionality one. +(1/10)! and (2/10)! are sufficient to express (N/10)! for all N. +(1/12)! and (2/12)! are sufficient to express (N/12)! for all N. +(1/3)! and (1/4)! are sufficient to express (N/12)! for all N. +Thus the three cases above are of dimensionality two. +PROBLEM: Find some order to this dimensionality business. +The reflection and multiplication formulas: + Z!(-Z)! = pi*Z/SIN pi*Z + + (N-1)/2 -NZ-1/2 +(2*pi) N (NZ)! = Z!(Z-1/N)!(Z-2/N)! ... (Z-(N-1)/N)! + +ITEM (Jan Kok): +PROBLEM: Given a regular n-gon with all diagonals drawn, how many +regions are there? In particular, how many triple (or N-tuple) +concurrences of diagonals are there? + +ITEM (Schroeppel): +Regarding convergence of Newton's method for quadratic equations: +Draw the perpendicular bisector of the line connecting the two +roots. Points on either side converge to the closest root. +On the line: +1 they do not converge +2 there is a dense set of points which involve division by zero +3 there is a dense set of points which loop, but roundoff error + propagates so all loops are unstable +4 being on the line is also unstable (if the roots are imaginary +and you are on the real axis, you may be doing exact computation +of the imaginary part (0), hence stay on the line. Example: +X^2 + 1 = 0, X0 = random real floating point number.) + ITEM (Schroeppel): +By Mathlab, the discriminant of X^4 + FX^3 + GX^2 + HX + I is +(as the discriminant of AX^2 + BX + C is B^2 - 4AC): + - 27 H^4 + 18 FGH^3 - 4 F^3H^3 - 4 G^3H^2 + F^2G^2H^2 + + I * [144 GH^2 - 6 F^2H^2 - 80 FG^2H + + 18 F^3GH + 16 G^4 - 4 F^2G^3] + + I^2 * [- 192 FH - 128 G^2 + 144 F^2G - 27 F^4] + - 256 I^3 + +ITEM: +In general, the discriminant of an n-th degree polynomial is +PRODUCT(ROOT - ROOT )^2 = square of determinant whose i,j element + i N +f(I;X,Y,...) = 1 if I = 0 + X*f(I-1;Y,Z,...) + f(I;Y,Z,...) (N-1 variables) + +The generating function is simply + N I +sum f(I;X,Y,Z,...)*S = (1+XS)(1+YS)(1+ZS)... +I=0 + +ITEM (Schroeppel): +Solutions to F(X) = X^3 - 3BX^2 + CX + D = 0 are +B - K * CUBE ROOT [F(B)/2 + SQRT [(F(B)/2)^2 + (F'(B)/3)^3]] + - K^2 * CUBE ROOT [F(B)/2 - SQRT [(F(B)/2)^2 + (F'(B)/3)^3]] +where K is one of the three cuberoots of 1: +1, (-1+SQRT(-3))/2, (-1-SQRT(-3))/2. + +ITEM (Schroeppel & Salamin): +If X^4 + BX^2 + CX + D = 0, then 2X = (SQRT Z1) + (SQRT Z2) + (SQRT Z3) +Z1, Z2, Z3 are roots of Z^3 + 2BZ^2 + (B^2 - 4D)Z - C^2 = 0. +The choices of square roots must satisfy (SQRT Z1)(SQRT Z2)(SQRT Z3) = -C. + ITEM (Salamin): +An easy solution of -4X^3 + 3X - a = 0 is X = sin((arcsin a)/3). +In a similar manner, the general quintic can be solved exactly by +use of the elliptic modular function and its inverse. +See Davis: Intro. to Nonlinear Differential and Integral +Equations (Dover), p. 172. Unfortunately, there exists +>= 1 typo, since his eqs. (7) and (13) are inconsistent. + +ITEM (Salamin): +The following operations generate one-to-one conformal mappings +of Euclidean N-space onto itself. + 1) Translate N-space. + 2) Expand N-space about one of its points. + 3) Stereographically project N-space onto an N-sphere, + rotate the sphere, then project back onto N-space. +PROBLEMS: +Show that all such conformal maps are generated by these +operations for any N. If the one-to-one and onto conditions are +removed, then for N=2, conformal maps can be obtained by analytic +functions. Show that for N>2, no new conformal maps exist. + ************************************************** +RECURRENCE RELATIONS +************************************************** + +ITEM (Gosper & Salamin): +"the Fast Fibonacci Transform" (motivation for next item) +Define multiplication on ordered pairs + (A,B)(C,D)=(AC+AD+BC,AC+BD). +This is just (AX+B)*(CX+D) mod X^2-X-1, and so is associative, +etc. We note (A,B)(1,0)=(A+B,A), which is the Fibonacci + N +iteration. Thus, (1,0) = (FIB(N),FIB(N-1)), which can be +computed in LOG N steps by repeated squaring, for instance. +FIB(15) is best computed using N=16, thus pushing the minimal +binary addition chain counterexample to 30 (Liknaitzky). (See +Knuth vol. 2, p 398.) By the last formula, + -1 + (1,0) = (FIB(-1),FIB(-2)) = (1,-1), +which, as a multiplier, backs up one Fibonacci step (further +complicating the addition chain question). Observing that + (1,0)^0 = (FIB(0),FIB(-1)) = (0,1) += the (multiplicative) identity, equate it with scalar 1. Define +addition and scalar multiplication as with ordinary vectors. + -1 + (A,B) = (-A,A+B)/(B^2+AB-A^2), +so we can compute rational functions when the denominator isn't +zero. Now, by using power series and Newton's method, we can + (X,Y) +compute fractional Fibonaccis, and even e and LOG(X,Y). +If we start with (1,0) and square iteratively, the ratio will +converge to the larger root of x@2-x-1 (= the golden ratio) +about as rapidly as with Newton's method. This +method generalizes for other polynomial roots, being an +improvement of the method of Bernoulli and Whittaker (Rektorys, +Survey of Applicable Math., p 1172). For the general second +order recurrence, F(N+1) = XF(N) + YF(N-1), we have the +multiplication rule: (A,B)(C,D) = (AD+BC+XAC,BD+YAC). + -1 +Inverse: (A,B) = (-A,XA+B)/(B^2+XAB-YA^2). + N +Two for the price of one: (F(1),YF(0))(1,0) = (F(N+1),YF(N)). + ITEM (Salamin & Gosper): + LINEAR RECURRENCE RELATIONS +Recurrence relation: A = C A + ... + C A (1) + k n-1 k-1 0 k-n + +with A , ... , A given as initial values. + 0 n-1 +Consider the algebra with basis vectors + n-1 + X^0, X^1, X^2, ... , X + n n-1 +and the identification X = C X + ... + C X^0. (2) + n-1 0 +Thus if U, V, W are vectors and W = U V, then componentwise + W = SUM T U V , (3) + i j,k ijk j k + +where the T's depend only on the C's. The following simple + k +procedure yields A : express the vector X as a linear + k m +combination of the basis vectors, then set X = A (0 =< m < n). + k m +Computation of X can be done by k-n+1 applications of (2) or by +computing the T's in (3) and then applying (3) O(log k) times. + k +PROOF: If 0 <= k < n, X is already a basis vector, so we get A . + L n L-n k +Suppose the procedure works for k < L. X = X X + n-1 L-n + = (C X + ... + C ) X + n-1 0 + L-1 L-n + = C X + ... + C X + n-1 0 + m L +The procedure evaluates each X to A , so X evaluates to + m + C A + ... + C A = A . QED + n-1 L-1 0 L-n L +The same procedure will work for negative k using + -1 n-1 n-2 + X = (X - C X - ... - C )/C , (4) + n-1 1 0 +the unique vector which when multiplied by X yields X^0. + Let (2) be F(X)=0 and V be the algebra constructed above. +Then V is a field iff F(X) is irreducible in the field of the +coefficients of V. +PROOF: Note that an element P of V is zero iff P(X)=0 mod F(X). +If G(X) H(X)=F(X), DEG G,H < DEG F, then the product of two +non-zero elements is zero and so V can't be a field. + Let P be an arbitrary non-zero element of V. +DEG(GCD(P,F)) <= DEG P < DEG F. +If F(X) is irreducible, then GCD(P,F)=1, so there exist +Q(X), R(X) such that Q(X) P(X) + R(X) F(X) = 1. +Then Q(X) P(X)=1 mod F(X). Since P has an inverse, V is a field. + +ITEM (Gosper & Salamin): +Yet another way to rapidly evaluate recurrences is to observe +that if F(N+1) = X*F(N) + Y*F(N-1), +then F(N+2) = (X^2+2Y)*F(N) - Y^2*F(N-2). +This rate doubling formula can be applied iteratively to compute +the Nth term in about LOG N steps, e.g., to get the 69th term +given terms 1 and 2, we form 1, 2, 3, 5, 9, 13, 21, 37, 69. +This sequence is computed from right to left by iteratively +subtracting half the largest possible power of 2. This is +sufficient to guarantee that some term further left will differ +from the left one by that same (halved) power of 2; e.g., 5, +...,21,37 have a common difference of 2^4, so that term 37 can be +found from term 5 and term 21 using the fourth application of the +rate doubling formula. + +The rate tripling formula is F(N+3) = (X^3+3XY)*F(N) + Y^3*F(N-3). +For the K-tupling formula: F(N+K) = P(K)*F(N) + Q(K)*F(N-K) +P(K+1) = X*P(K) + Y*P(K-1) (the same recurrence as F) +Q(K+1) = -Y*Q(K) +P(1) = X Q(1) = Y +P(0) = 2 Q(0) = -1 + K +Q(K) = -(-Y) + K/2 +P(K) = 2(-Y) *T(K;X/SQRT(-4Y)) where T(K;X) is the +Kth Chebychev polynomial = cos (K arccos X) + +If A(I), B(I), and C(I) obey the same second order recurrence, + +( A B )-1 ( C ) +( I I ) ( I ) +( ) ( ) (I) +( A B ) ( C ) +( J J ) ( J ) + +is independent of I and J, provided the inverse exists. +(This is true even if coefficients are not constant, +since any two independent sequences form a basis.) + Plugging in F and P as defined above, we get an expression for +the Nth term of the general second order recurrence in terms of +P(N) and P(N+1): + +(P(N) P(N+1)) ( P(0) P(1) )-1 ( F(0) ) + ( P(1) P(2) ) ( F(1) ) = F(N). + +Setting X = Y = 1, we get FIB(N) = (2P(N+1)-P(N))/5, which is a +complex but otherwise square root free closed form. (SQRT(-4) = 2i) + +With constant coefficients, the invariance (I) implies: + +(A A ) ( A A )-1 ( A ) + P+I P+J ( Q+I+K Q+J+K ) ( R+K ) + ( ) ( ) = A + ( A A ) ( A ) P-Q+R + ( Q+I+L Q+J+L ) ( R+L ) + +These matrix relations generalize directly +for Nth order recurrences. + +ITEM (Chebychev): +The Nth Chebychev polynomial T(N) = T(N;x) = cos (N arccos x). +T(0) = 1, T(1) = x, T(N+1) = 2x T(N) - T(N-1). + N 1-N +T(N;T(M)) clearly = T(NM). x - 2 T(N), whose degree is N-2, + N +is the polynomial of degree < N which stays closest to x in the + 1-N +interval (-1,1), deviating by at most 2 at the N+1 places +where x = cos(K*pi/N), K=0,1,...N. + N +Generating function: SUM T(N)*S = (1-xS)/(1-2xS+S^2). +First order (nonlinear) recurrence: + T(N+1) = xT(N) - SQRT((1-x^2)(1-T(N)^2)). + N +(T(N+1),-T(N)) = (T(1),-T(0))(1,0) , + where (A,B)(C,D) = (AD+BC+2xAC,BD-AC). + +ITEM: n n + 1 (1+ix) - (1-ix) +tan (n arctan x) = - * ----------------- + n n + i (1+ix) + (1-ix) + ********************************************* +BOOLEAN ALGEBRA +********************************************* + +ITEM (Schroeppel): +Problem: Synthesize a given logic function or set of functions +using the minimum number of two-input AND gates. NOT gates are assumed +free. Feedback is not allowed. The given functions are allowed to +have X (don't care) entries for some values of the variables. +P XOR Q requires three AND gates. MAJORITY(P,Q,R) requires 4 AND gates. +"PQRS is a prime number" seems to need seven gates. +The hope is that the best Boolean networks for functions might lead to +the best algorithms. + +ITEM (Speciner): + number of monotonic increasing Boolean +N functions of N variables +0 2 (T, F) +1 3 (T, F, P) +2 6 (T, F, P, Q, P AND Q, P OR Q) +3 20 +4 168 = 8 * 3 * 7 +5 7581 = 3 * 7 * 19^2 +6 7,828,354 = 2 * 359 * 10903 (Ouch!) +N from 0 to 4 suggest that a formula should exist, but 5 and 6 +are discouraging. A difficult generalization: Given two partial +orderings, find the number of maps from one to the other that are +compatible with the ordering. A related puzzle: A partition of +N is a finite string of non-increasing integers that add up to N. +Thus 7 3 3 2 1 1 1 is a partition of 18. Sometimes an infinite +string of zeros is extended to the right, filling a half-line. +The number of partitions of N, P(N), is a fairly well understood +function. inf n inf k +The generating function is sum P(n) x = 1 / prod (1-x ) . + n=0 k=1 +A planar partition is like a partition, but the entries are in a +two-dimensional array (the first quadrant) instead of a string. +Entries must be non-increasing in both the x and y directions. +A planar partition of 34 would be: 1 + 3 1 + 3 2 2 1 + 7 6 4 3 1 +Zeros fill out the unused portion of the quadrant. The number of +planar partitions of n, PL(n), is not a very well understood +function. inf n inf k k +The generating function is sum PL(n) x = 1 / prod (1-x ) . + n=0 k=1 +No simple proof of the generating function is known. Similarly, +one can define cubic partitions with entries in the first octant, +but no one has been able to discover the generating function. +Some counts for cubic partitions and a discussion appear in +Knuth, Math. Comp. 1970 or so. + ITEM (Schroeppel): +The 2-NOTs problem: Synthesize a black box which computes NOT-A, +NOT-B, and NOT-C from A, B, and C, using an arbitrary number of +ANDs and ORs, but only 2 NOTs. + +Clue: (Stop! Perhaps you would like to work on this awhile.) +Lemma: Functions synthesizable with one NOT are those where +the image of any upward path (through variable space) has at +most one decrease (that is, from T to F). + +ITEM (Roger Banks): +A Venn diagram for N variables where the shape representing each +variable is convex can be made by superimposing successive M-gons +(M = 2, 4, 8, ...), every other side of which has been pushed out +to the circumscribing circle. If you object to superimposed +boundaries, you may shrink the nested M-gons a very slight amount +which depends on N. + +ITEM (Schroeppel & Waltz): +PROBLEM: Cover the Execuport character raster completely with the +minimum number of characters. The three characters I, H and # +works. Using capital letters only, the five characters B, I, M, +V and X is a minimal solution. Find a general method of solving +such problems. + +ITEM (Gosper): +PROBLEM: Given several binary numbers, how can one find a mask +with a minimal number of 1 bits, which when AND-ed with each of +the original numbers preserves their distinctness from each +other? What about permuting bit positions for minimum numerical +spread, then taking the low several bits? + +ITEM (Schroeppel): +(A AND B) + (A OR B) = A + B = (A XOR B) + 2(A AND B). + +ITEM (Minsky): +There exists a convex figure n congruent copies of which, +for any n, form a Venn diagram of 2^n regions. + ********************************************* +RANDOM NUMBERS +********************************************* + +ITEM (Schroeppel): +Random number generators, such as Rollo Silver's favorite, +which use SHIFTs and XORs, and give as values only some part +of their internal state, can be inverted. Also, the outputs +may often be used to obtain their total internal state. +For example, 2 consecutive values from Rollo's suffice +to allow prediction of its entire future. Rollo's is: +RANDOM: MOVE A,HI ;register A gets loaded with "high" word + MOVE B,LO ;register B gets loaded with "low" word + MOVEM A,LO ;register A gets stored in "low" word + LSHC A,35. ;shift the 72-bit register AB left 35 + XORB A,HI ;bitwise exclusive-or of A and HI + replaces both +This suggests a susceptibility to +analysis of mechanical code machines. + +See LOOP DETECTOR item in FLOWS AND ITERATED FUNCTIONS section. + +ITEM (? via Salamin): +A mathematically exact method of generating a Gaussian +distribution from a uniform distribution: let x be uniform on +[0,1] and y uniform on [0,2 pi], x and y independent. Calculate +r = SQRT(-log x). Then r cos y and r sin y are two independent +Gaussian distributed random numbers. + +ITEM (Salamin): +PROBLEM: Generate random unit vectors in N-space uniform on the +unit sphere. SOLUTION: Generate N Gaussian random numbers and +normalize to unit length. + ********************************************* +NUMBER THEORY, PRIMES, PROBABILITY +********************************************* + +ITEM (Schroeppel): +After about 40 minutes of run time to verify the absence of +any non-trivial factors less than 2^35, the 125th Mersenne +number, 2^125 - 1, was factored on Tuesday, January 5, 1971, +in 371 seconds run time as follows: 2^125 - 1 = +31 * 601 * 1801 * 26 90898 06001 * 4710 88316 88795 06001. +John Brillhart at the University of Arizona had already +done this. M137 was factored on Friday, July 9, 1971 in +about 50 hours of computer time: 2^137 - 1 = +32032 21559 64964 35569 * 54 39042 18360 02042 90159. + +ITEM (Schroeppel): +For a random number X, the probability of its largest prime +factor being (1) greater than the square root of X is LN 2. +(2) less than the cube root of X is about 4.86%. This suggests +that similar probabilities are independent of X; for instance, +the probability that the largest prime factor of X is less than +the 20th root of X may be a fraction independent of the size of X. +RELEVANT DATA: +([ ] denote the expected value of adjacent entries.) + +RANGE COUNT CUMULATIVE SUM OF COUNT +10^12 TO 10^6 7198 [6944] 10018 +10^6 TO 10^4 2466 2820 +10^4 TO 10^3 354 402 [487] 2.4 +10^3 TO 252 40 48 ;252 = 10 +252 TO 100 7 8 1.7 +100 TO 52 1 1 ;52 = 10 +51 TO 1 0 0 +where: +"COUNT" is the number of numbers between 10^12 + 1 and +10^12 + 10018 whose largest prime factor is in "RANGE". +The number of primes in 10^12 + 1 to 10^12 + 10018 is 335; the +prime number theorem predicts 363 in this range. This is +relevant to Knuth's discussion of Legendre's factoring method, +vol. 2, p. 351-354. + +ITEM (Schroeppel): +Twin primes: + 166,666,666,667 = (10^12 + 2)/6 + 166,666,666,669 +The number 166,666,666,666,667 is prime, but + 166,666,666,666,669 is not. +The primes which bracket 10^12 are 10^12 + 39 and 10^12 - 11. +The primes which bracket 10^15 are 10^15 + 37 and 10^15 - 11. +The number 23,333,333,333 is prime. +Various primes, using T = 10^12, are: +40T + 1, 62.5T + 1, 200T - 3, 500T - 1, 500T - 7. + +ITEM (Fredkin): + 3 3 3 3 +3 + 4 + 5 = 6 . + ITEM (Schroeppel): +91038 90995 89338 00226 07743 74008 17871 09376^2 = + 82880 83126 51085 58711 66119 71699 91017 17324 + 91038 90995 89338 00226 07743 74008 17871 09376 + +ITEM (Schroeppel): N +Ramanujan's problem of solutions to 2 - 7 = X^2 was searched +to about N = 10^40; only his solutions (N = 3, 4, 5, 7, 15) were +found. It has recently been proven that these are the only ones. +Another Ramanujan problem: Find all solutions of n! + 1 = x^2. + +ITEM (Schroeppel): +Take a random real number and raise it to large powers; we expect +the fraction part to be uniformly distributed. Some exceptions: +1 -- PHI = (1 + SQRT 5)/2 +2 -- all -1 < X < 1 +3 -- SQRT 2 (half are integers, other half + are probably uniformly distributed) +4 -- 1 + SQRT 2 -- Proof: + N N + (1 + SQRT 2) + (1 - SQRT 2) = integer (by induction); + N + the (1 - SQRT 2) goes to zero. +5 -- 2 + SQRT 2 -- similar to 1 + SQRT 2 +6 -- any algebraic number whose conjugates + are all inside the unit circle +Now, 3 + SQRT 2 is suspicious; it looks non-uniform, and seems to +have a cluster point at zero. PROBLEM: Is it non-uniform? + +ITEM (Schroeppel): +Numbers whose right digit can be repeatedly removed and +they are still prime: CONJECTURE: There are a finite +number of them in any radix. In decimal there are 51, +the longest being 1,979,339,333 and 1,979,339,339. + +ITEM (Schroeppel): +PROBLEM: Can every positive integer be expressed +in terms of 3 and the operations factorial and +integer square root? E.g., 5 = SQRT(SQRT 3!!). + +ITEM (Schroeppel): +Take as many numbers as possible from 1 to N such that no 3 are +in arithmetic progression. CONJECTURE: As N approaches infinity, the + (LN 2)/(LN 3) +density of such sets approaches zero, probably like N . + XX.XX is a known solution for N = 5 + XX.XX....XX.XX is a known solution for N = 14 +Conjecture that XX.XX just keeps getting copied. If the + (LN 2)/(LN 3) +N can be proved, it follows that there are +infinitely many primes P1, P2, P3 in arithmetic progression, + (LN 2)/(LN 3) +since primes are much more common than N . + +ITEM (Schroeppel): +PROBLEM: How many squares have no zeros +in their decimal expression? Ternary? + ITEM (Gosper): +The number of n digit strings base B in which all B digits occur +at least once is just the Bth forward difference at 0 of the nth +powers of 0, 1, 2, ... . Eg., for n = 4: + 0 1 16 81 256 625 + 1 15 65 175 369 + 14 50 110 194 + 36 60 84 + 24 24 + 0 + +so there are 14 (= 2^4-2) such 4-bit strings, 36 such 4-digit +ternary strings, 24 (= 4!) such quaternary, and 0 for all higher +bases. 27 (= 10e?) random decimal digits are required before it +is more likely than not that every digit has occurred; with 50 +digits the likelihood is 95%. + +ITEM (Fredkin): +By the binomial theorem, the bth forward difference at 0 of +the 0, 1, 2, ... powers of n is (n-1)^b. E.g., for n = 4: + 1 4 16 64 256 + 3 12 48 192 + 9 36 144 + 27 108 + 81 +In fact, any straight line with rational slope through such an +array will always go through a geometric sequence with common +ratio of the form n^a (n-1)^b. In the above, east by southeast +knight's moves give the powers of 12: 1, 12, 144, ... . + +ITEM (Schroeppel, etc.): +The joys of 239 are as follows: +Pi = 16 ARCTAN (1/5) - 4 ARCTAN (1/239), +which is related to the fact that 2 * 13^4 - 1 = 239^2, +which is why 239/169 is an approximant (the 7th) of SQRT 2. +ARCTAN (1/239) = ARCTAN (1/70) - ARCTAN (1/99) + = ARCTAN (1/408) + ARCTAN (1/577) +239 needs 4 squares (the maximum) to express it. +239 needs 9 cubes (the maximum, shared only with 23) to express it. +239 needs 19 fourth powers (the maximum) to express it. +(Although 239 doesn't need the maximum number of fifth powers.) +1/239 = .00418410041841..., which is related to the fact that +1,111,111 = 239 * 4,649. +The 239th Mersenne number, 2^239 - 1, is known composite, + but no factors are known. +239 = 11101111 base 2. +239 = 22212 base 3. +239 = 3233 base 4. +There are 239 primes < 1500. +K239 is Mozart's only work for 2 orchestras. +Guess what memo this is. +And 239 is prime, of course. + ITEM (Salamin): +There are exactly 23000 primes less than 2^18. + +ITEM (Gosper): +To show that N+1 L N+1 L +SUM (L=0 to N) (BINOMIAL N+L L)*(X (1-X) + (1-X) X ) = 1 +set N to 20 and observe that it is the probability that one or +the other player wins at pingpong. (X = probability of first +player gaining one point, L = loser's score, deuce rule +irrelevant.) If this seems silly, try more conventional methods. +PROBLEM: If somehow you determine A should spot B 6 points for +their probabilities of winning to be equal, and B should spot C +9 points, how much should A give C? + +ITEM (Schroeppel): +Let (A,B,C...) be the multinomial coefficient +(A+B+C...)!/A!B!C!... +This is equal modulo the prime p to +(A0,B0,C0...)(A1,B1,C1...)(A2,B2,C2...)... +where AJ is the Jth from the right digit of A base p. +Thus (BINOMIAL A+B A) mod 2 is 0 iff (AND A B) is not. +The exponent of the largest power of p which divides (A,B,C...) +is equal to the sum of all the carries when the base p expressions +for A, B, C, ... are added up. + +ITEM (Gosper): +Recurrences for multinomial coefficients: +(A,B,C,...) = (A+B,C,...)(A,B) = (A+B+C,...)(A,B,C) = ... + +PROBLEM (Gosper): +Take a unit step at some heading (angle). +Double the angle, step again. Redouble, step, etc. +For what initial heading angles is your locus bounded? + +PARTIAL ANSWER (Schroeppel, Gosper): When the initial angle is +a rational multiple of pi, it seems that your locus is bounded +(in fact, eventually periodic) iff the denominator contains as a +factor the square of an odd prime other than 1093 and 3511, which +must occur at least cubed. (This is related to the fact that +1093 and 3511 are the only known primes satisfying + P 2 +2 = 2 mod P ). But a denominator of 171 = 9 * 19 never loops, +probably because 9 divides PHI(19). Similarly for 9009 and 2525. +Can someone construct an irrational multiple of pi with a bounded +locus? Do such angles form a set of measure zero in the reals, +even though the "measure" in the rationals is about .155? +About .155 = the fraction of rationals with denominators +containing odd primes squared = 1 - PRODUCT over odd primes of +1 - 1/P(P + 1). This product = .84533064 +or- a smidgen, and +is not, alas, SQRT(pi/2) ARCERF(1/4) = .84534756. This errs by 16 +times the correction factor one expects for 1093 and 3511, and is +not even salvaged by the hypothesis that all primes > a million +satisfy the congruence. It might, however, be salvaged by +quantities like 171. + ITEM (Schroeppel): +The most probable suit distribution in bridge hands is 4-4-3-2, +as compared to 4-3-3-3, which is the most evenly distributed. +This is because the world likes to have unequal numbers: +a thermodynamic effect saying things will not be in the state of +lowest energy, but in the state of lowest disordered energy. + +ITEM (Beeler): +The Fibonacci series modulo P has been studied. This series has +a cycle length L and within this cycle has sub-cycles which are +bounded by zero members. +The length of powers of primes seems to be + power-1 + L = (length of prime) * prime +The length of products of powers of primes seems to be + L = least common multiple of lengths of powers of primes + which are factors. +There can be only 1, 2 or 4 sub-cycles in the cycle of a prime. +Primes with 1 sub-cycle seem to have lengths + L = (prime - 1)/N, N covering all integers. +Primes with 2 sub-cycles seem to have lengths + M + L = (prime - (-1) )/M, M covering + all integers except of form 10 K + 5. +Primes with 4 sub-cycles seem to always be of form 4 K + 1, +and seem to have lengths + L = 2 (prime + 1)/R or (prime - 1)/S, + R covering all integers of form 10 K + 1, 3, 7 or 9; + S covering all integers. +At Schroeppel's suggestion, the primes have been separated +mod 40, which usually determines their number of sub-cycles: +PRIME mod 40 SUB-CYCLES +1, 9 usually 2, occasionally 1 or 4 (about equally) +3, 7, 23, 27 2 +11, 19, 31, 39 1 +13, 17, 33, 37 4 +21, 29 1 or 4 (about equally) +2 (only 2) 1 +5 (only 5) 4 +Attention was directed to primes which are 1 or 9 mod 40 but have +1 or 4 subcycles. 25 X^2 + 16 Y^2 seems to express those which are +9 mod 40; (10 X + or - 1)^2 + 400 Y^2 seems to express those which +are 1 mod 40. PROBLEM: Can some of the "seems" above be proved? +Also, can a general test be made which will predict exact length +for any number? + +ITEM (Gosper, Schroeppel): +A point of the 2 dimensional lattice is called visible iff its +coordinates are relatively prime. The invisible 2 by 2 square +with smallest X has its near corner on (14,20). +(I.e., (14,20), (15,20), (14,21), and (15,21) are all invisible.) +The corresponding 3 by 3 is at (104,6200). By the Chinese +remainder theorem, there exist invisible sets of every finite +shape. Excellent reference: Amer. Math. Monthly, May '71, p487. + ITEM : +There is a unique "magic hexagon" of side 3: + 3 17 18 First discovered by Clifford W. Adams, + 19 7 1 11 who worked on the problem from 1910. +16 2 5 6 9 In 1957, he found a solution. + 12 4 8 14 (See Aug. 1963 Sci. Am., Math. Games.) + 10 13 15 Other length sides are impossible. + +ITEM (Schroeppel): +There is no magic cube of order 4. +Proof: Let K (= 130) be the sum of a row. +Lemma 1: In a magic square of order +four, the sum of the corners is K. +Proof: Add together each edge of the square and the 2 diagonals. +This covers the square entirely, and each corner twice again. +This adds to 6K, so twice the corner sum is 2K. +Lemma 2: In a magic cube of order 4, the sum of any +two corners connected by an edge of the cube is K/2. +Proof: Call the corners a and b. Let c, d and e, f be the +corners of any two edges of the cube parallel to ab. Then +abcd, abef, and cdef are all the corners of magic squares. So +a+b+c+d + a+b+e+f + c+d+e+f = 3K; a+b+c+d+e+f = 3K/2; a+b = K/2. +Proof of magic cube impossibility: Consider a corner x. +There are three corners connected by an edge to x. +Each must have value K/2 - x. QED + +ITEM (Schroeppel): +By similar reasoning, the center of an order 5 magic cube must +be 63 = K/5. COROLLARY: There is no magic tesseract of order 5. + +ITEM (Salamin): +The probability that two random integers are relatively prime is +6/pi^2. PSEUDO-PROOF: Let X be the probability. Let S be the set of +points in the integer lattice whose coordinates are relatively prime, so +that S occupies a fraction X of the lattice points. Let S(D) be +the set of points whose coordinates have a GCD of D. S(D) is S +expanded by a factor of D from the origin. So S(D) occupies a +fraction X/D^2 of the lattice, or the probability that two random +integers have a GCD of D is X/D^2. If D unequals D', then S(D) +intersect S(D') is empty, and union of all S(D) is the entire +lattice. Therefore X*(1/1^2+1/2^2+1/3^2+...) = 1, so X = 6/pi^2. +[This argument is not rigorous, but can be made so.] + +ITEM (Salamin): +The probability that N numbers will lack +a Pth power common divisor is 1/ZETA(NP). + +ITEM (Salamin & Gosper): +The probability that a random rational number +has an even denominator is 1/3. + +ITEM (Schroeppel): GAUSSIAN INTEGERS +See following illustrations; also PI section. + ITEM 56 (Beeler): page +The "length" of an N-digit decimal number is defined as the +number of times one must iteratively form the product of its +digits until one obtains a one-digit product (see Technology +Review Puzzle Corner, December 1969 and April 1970). For various +N, the following shows the maximum "length", as well as how many +distinct numbers (permutation groups of N digits) there are: +N MAX L DISTINCT + 2 4 54 + 3 5 219 + 4 6 714 + 5 7 2,001 + 6 7 5,004 + 7 8 11,439 + 8 9 24,309 + 9 9 48,619 +10 10 92,377 +11 10 167,959 +12 10 293,929 +Also, for N = 10, 11 and 12, a tendency for there to be many +fewer numbers of "length" = 7 is noted. Other than this, the +frequency of numbers of any given N, through N = 12, decreases +with increasing "length". CONJECTURE (Schroeppel): No L > 10. + +ITEM 57 (Beeler, Gosper): +There is at least one zero in the decimal expression of each +power of 2 between 2@8@6 = 77,371,252,455,336,267,181,195,264 +and 2@3@0@7@3@9@0@1@4, where the program was stopped. If digits of +such powers were random, the probability that there is another +zeroless power would be about 1/10@4@1@1@8@1@6. Assuming there aren't +any then raises the question: +How many final nonzero digits can a power of two have? +ANSWER (Schroeppel): Arbitrarily many. If we look at the last +n digits of consecutive powers of 2, we see: +a) None end in zero. + n +b) After the nth, they are all multiples of 2 . + n-1 +c) They get into a loop of length 4 * 5 . + (Because 2 is a primitive root of powers of 5.) + n-1 n +But there are only 4 * 5 multiples of 2 which don't end with + n +zero and are < 10 , so we will see them all. In particular, we +will see the one composed entirely of 1's and 2's, which ends +...11112111211111212122112. +ANSWER (Schroeppel): Arbitrarily many. If we look at the last +n digits of consecutive powers of 2, we see: +a) None end in zero. + n +b) After the nth, they are all multiples of 2 . + n-1 +c) They get into a loop of length 4 * 5 . + (Because 2 is a primitive root of powers of 5.) + n-1 n +But there are only 4 * 5 multiples of 2 which don't end with + n +zero and are < 10 , so we will see them all. In particular, we +will see the one composed entirely of 1's and 2's, which ends +...11112111211111212122112. + ITEM (Beeler): +If S = the sum of all integers which exactly divide N, +including 1 and N, then "perfect numbers" are S = 2 N; +the first three numbers which are S = 3 N are: +120 = 2^3 * 3 * 5 = 1111000 base 2 +672 = 2^5 * 3 * 7 = 1010100000 base 2 +523,776 = 2^9 * 3 * 11 * 31 = 1111111111000000000 base 2 + +ITEM (Root): +Consider iteratively forming the sum of the factors (including 1 +but not N) of a number N. This process may loop; "perfect +numbers" are those whose loop is one member, N. For example, +N = 28 = 1 + 2 + 4 + 7 + 14. An example of a two-member loop is: + sum of factors of 220 = 284 + sum of factors of 284 = 220 +Two-member loops are called "amicable pairs." +A program to search for loops of length > 2, all of whose +members are < 6,600,000,000 found the known loops of length 5 +(lowest member is 12496) and 28 (lowest member is 14316), +but also 13 loops of 4 members (lowest member is given): + 1,264,460 = 2^2 * 5 * 17 * 3,719 + 2,115,324 = 2^2 * 3^2 * 67 * 877 + 2,784,580 = 2^2 * 5 * 29 * 4,801 + 4,938,136 = 2^3 * 7 * 109 * 809 + 7,169,104 = 2^4 * 17 * 26,357 + 18,048,976 = 2^4 * 11 * 102,551 + 18,656,380 = 2^2 * 5 * 932,819 + 46,722,700 = 2^2 * 5^2 * 47 * 9,941 + 81,128,632 = 2^3 * 13 * 19 * 41,057 + 174,277,820 = 2^2 * 5 * 29 * 487 * 617 + 209,524,210 = 2 * 5 * 7 * 19 * 263 * 599 + 330,003,580 = 2^2 * 5 * 16,500,179 + 498,215,416 = 2^3 * 19 * 47 * 69,739 + +ITEM (Schutzenberger): +PROBLEM: Using N digits, construct a string of digits which +at no time has any segment appearing consecutively twice. + N = 2 => finite maximum string + N = 10 => known infinite +Determine maximum string length for N = 3. +SUB-PROBLEM: How many sequences exist of any particular length? + +ITEM (Gosper): +The variance of a pseudo-Gaussian distributed random variable +made by adding T independent, uniformly distributed random +integer variables which range from 0 to N-1, inclusive, is +T((N^2 - 1)/12). + ITEM (Speciner): +The first four perfect numbers are 6, 28, 496, 8128. +Two-member loops (amicable pairs) are: + 220 @E@T 284 + 1184 @E@T 1210 + 2620 @E@T 2924 + 5020 @E@T 5564 + 6232 @E@T 6368 + 10744 @E@T 10856 + 12285 @E@T 14595 + 17296 @E@T 18416 + 63020 @E@T 76084 + 66928 @E@T 66992 + 67095 @E@T 71145 + 69615 @E@T 87633 + 79750 @E@T 88730 + 100485 @E@T 124155 + 122265 @E@T 139815 + 122368 @E@T 123152 + 141644 @E@T 153176 + 142310 @E@T 168730 + 171856 @E@T 176336 + 176272 @E@T 180848 + 185368 @E@T 203432 + 196724 @E@T 202444 +(Exhaustive to smaller member =< 196724 and larger member < 2^35.) +A prime decade is where N+1, N+3, N+7 and N+9 are all prime. +The first occurrence of two prime decades with the theoretical +minimum separation is N = 1006300 and N = 1006330. The 335th +prime decade is N = 2342770. There are 172400 primes < 2342770. + ********************************************* +AUTOMATA THEORY +********************************************* + +ITEM (Schroeppel): +A 2-counter machine, given N in one of the counters, cannot + N +generate 2 . Proven Saturday, September 26, 1970. (Independently +rediscovered by Frances Yao.) But (Minsky, Liknaitzky), given + N + N 2 +2 , it can generate 2 . (A 2-counter machine has a fixed, +finite program containing only the instructions "ADD 1", +"SUBTRACT 1", "JUMP IF NOT ZERO", which refer to either of two +unlimited counters. Such machines are known universal, but +(due to the above) they must have specially encoded inputs.) + +ITEM (Schroeppel): +What effort is required to compute PI(X), .7 +the number of primes < X? Shanks and Brillhart claim about X . + +ITEM (Gosper): +See space-filling curve machine item in TOPOLOGY section. + ********************************************* +GAMES +********************************************* + +ITEM (Schroeppel): +Regarding "poker coins" game, whose rules are: +1 a player throws N coins; + he then puts one or more aside and rethrows the rest +2 this throwing is repeated until he no longer has any to throw +3 highest score (dice) or maximum number of heads (coins) wins +For poker coins, the optimal strategy, with N coins thrown, is: +Z = number of zeros (tails) + if Z = 0, quit + if Z = 1, throw the zero + if 1 < Z < N, save one one, throw the other N-1 coins + if Z = N, save a zero, throw the other N-1 coins +The optimal strategy for poker dice is hairier. + +ITEM (Schroeppel): +PROBLEM: Solve Blackout, a game as follows: Two players alternate +placing X's on a rectangular grid. No two X's may appear +adjacent along a side or across the diagonal at a corner. +The last X wins. Some theory: The "indicator" for a position is: +make all possible moves from the given position. +Evaluate the indicator of each of these successor positions. +The indicator of the first position is the smallest number which +is not the indicator of a successor position. The indicator of +the null position is 0. The second player wins iff the indicator +is 0. Example of calculating an indicator for the 3 x 3 board: +There are 3 distinct moves possible -- corner, side, center. +Playing in the center leaves the null position, indicator 0. +Playing on the side leaves a 1 x 3 line, indicator 2. Playing in +the corner leaves a 3 x 3 L, indicator 3. The smallest number +not appearing in our list is 1, so the indicator of a 3 x 3 +square is 1. For two boards (not touching) played +simultaneously, the indicator is the XOR of the indicators for +the separate boards. For any position, the indicator is <= the +maximum game length. +PROBLEM: Find some non-exponential way to compute the indicator +of a given position. For lines, a period of 34 is entered after +the line is about 80 long. For Ls: if one leg is held fixed, the +indicator (as a function of the other leg) seems to become +periodic with period 34. The time to enter the period becomes +greater as the fixed leg increases. +On an odd X odd board, the 1st player wins. +On a 4 X N board, the 2nd player wins. +On a 6 X 6 board, the 1st player wins by playing +at the center of one quarter. +This indicator analysis is similar for many other +take-away games, such as Nim. + +ITEM: +Berlekamp of Bell Labs has done the 9 squares +(16 dots) Dots game; the 2nd player wins. + +ITEM: +A neat chess problem, swiped from "Chess for Fun and Chess for +Blood", by Edward Lasker: white: pawns at QN3 and KN7, knight at +QN4, bishop at KB7, king at QB2; black: pawn at QN3, king at QR6. +White mates in three moves. + ITEM (Beeler): +There is only one distinct solution to the commercial +"Instant Insanity" colored-faces cubes puzzle, which is +how it comes packed. (Independently discovered by Dave Plumer.) +Mike Paterson has discovered a clever way to solve the puzzle. + +ITEM (Beeler): +A window-dice game is as follows: +1 The player starts with each of nine windows open, + showing the digits 1 - 9. +2 Roll two dice. +3 Cover up any digits whose sum is the sum on the dice. +4 Iterate throwing and closing windows until the equality of sums + is impossible. +5 Your score is the total of closed windows (highest wins). +An optimum strategy has been tabulated. Usually it is best +to take the largest digits possible, but not always; +it also depends critically on the remaining numbers. + +ITEM (Beeler): +Sim is a game where two players alternately draw lines connecting +six dots. The first person to form a triangle in his color +loses. The second player can always win, and whether his first +move connects with the first player's first move doesn't matter; +from there on, however, the strategy branches to a relatively +gruesome degree. +PROBLEM: 6 dots is minimum to ensure no stalemate with 2 players; +how many dots are required with 3 players? + +ITEM (Beeler): +The 4 X 4 game of Nim, also known as Tactix, is a win +for the second player, who on his first move can reply +center-symmetrically unless the first player's first move +was B1 and B2 (analyzed on RLE PDP-1). + +ITEM (Beeler): +Triangular Hi-Q (or peg solitaire) is 15 pegs in a triangle. +One peg is removed, and thereafter pegs jump others, +which are removed. With pegs numbered 1 at the top, +2 and 3 in the next row, etc., +REMOVE CAN END WITH ONLY THE PEG +1 1, 7 = 10, 13 +2 2, 6, 11, 14 +4 3 = 12, 4, 9, 15 +5 13 +Removing only one, no way exists to get to either 1 + 11 + 15 +(tips) or 4 + 6 + 13 (centers of sides). Starting with peg 1 +removed, 3,016 positions are attainable (not turning board); the +sum of ways to get to each of these is 10,306. An example is: +remove peg 1, then jump as follows: 6, 13, 10, 1, 2, 11, 14/13, +6, 12/13, 15, 7/4, 13, 4; leaving peg 1. + ITEM (Gosper, Brown, Rayfield): +A 1963 PDP-1 computer program gave us some interesting data on +the traditional game of peg solitaire (33 holes in a cross shape). + + A B C + D E F + G H I J K L M + N P Q . S T U + V W X Y Z 1 2 + 3 4 5 + 6 7 8 + +From the starting position, complement the board. This is +the ending position. Now from the starting position, make +one move, then complement the board. This is a position +one move from a win. By induction, you can win from the +complement of any position you can reach. Thus every successful +game has a dual game whose positions are the complements of the +original ones. This debunks the heuristic of emptying the arms +of the cross first and then cleaning up the middle, because +there are just as many dual games which empty out the +middle first and then the arms! The program found one +counterintuitive win which at one point left the center nine +empty but had ten in the arms. + + . B . + D E . + . . . . . . . + . P . . . T U + V W . . . . . + . 4 . + . 7 . + +By dualizing and permuting a solution from the folklore, +we found a similar winning position with 20. +(T Q 4 R 1 L J H W Y M J) leaves: + + A B C + D E F + G H . . . L . + N . . . . . U + V W . . . 1 2 + 3 . 5 + 6 7 8 + +(then 8 V A C/B 2 6 G M F/K S 8 1 Y V 3 Q A H E). + Another useful observation is that the pegs and their +original hole positions fall into four equivalence classes +in which they stay throughout the game. Thus the four pegs +which can reach the center on the first move are the only +ones that ever can. Similarly, the peg jumped over on +the last move must be in one of the two eight-membered +classes which get reduced on the first move. The program's +main heuristic was to reduce the larger classes first. + + a b a + c d c + a b a b a b a + c d c . c d c + a b a b a b a + c d c + a b a + +With its heuristics disabled, the program simply scanned +lexicographically (left to right in the inner loop, then +top to bottom) for a peg which could move. At one point, there +is a peg which can move two ways; it chose west. Twelve moves +from the end it stopped and went into an exhaustive "tree search", +in which it found two basically different wins. (Try it +yourself.) + + . . . + . . . + . . . . K . . + . . Q . . . . + . . X Y Z 1 2 + 3 4 5 + 6 7 8 + ********************************************* +PROPOSED COMPUTER PROGRAMS, IN ORDER OF +INCREASING RUNNING TIME (Schroeppel) +********************************************* + +PROBLEM: Count the polyominos up to, say, order 20. +From Applied Combinatorial Mathematics, pages 201 and 213: +ORDER E. H. NOT ENCLOSING HOLES +1 1 1 +2 1 1 +3 2 2 +4 5 5 +5 12 12 +6 35 35 +7 108 107 +8 369 363 +9 1285 1248 +10 4655 4271 +11 17073 +12 63600 +13 238591 +14 901971 +15 3426576 +16 13079255 +17 50107911 +18 192622052 +The order 13 through 18 data is from Computers in Number Theory, +1971, Atkin & Birch, ed., Academic Press, which has not been +independently checked. It also gives bounds 3.72 < limit as +N goes to infinity of Nth root of number of polyominos of order N +(including those enclosing holes) < 4.5. Also an asymptotic +formula for the number of polyominos: + N -.98+or-.02 +4.06 * (N ) * constant. Polyominos may be constructed +in 3-space (Soma-like pieces) or higher dimensions; a curious +thought is into how many dimensions does the average, say, +20-omino extend? + +PROBLEM: Solve "minichess", chess played on a 5 X 5 board +where each side has lost the king's rook, knight, bishop, +and 3 pawns, and the opponents are shoved closer together +(1 empty row intervening, no double pawn moves). + +PROBLEM: Solve the tiger puzzle, a sliding block puzzle mentioned +in Scientific American February 1964, pages 122 - 130. + +PROBLEM: Find smallest squared square (a square composed +entirely of smaller, unequal squares). Smallest known has 24 +small squares (Martin Gardner's Scientific American Book, +vol. 2, page 206). See also the following two illustrations. +Recently, someone constructed a squared rectangle with sides +in the ratio 1:2. It contains 1353 squares. + PROBLEM: Count the magic squares of order 5. There are +about 320 million, not counting rotations and reflections. + +PROBLEM: List (that is, count) the semigroups of 7 elements; +also, the groups of 256 elements (estimated: 11000). + +PROBLEM (Gosper): Compute the integer-valued step function F(R), +0= 1 except perhaps the first. This means that z' will always +exceed 1 and thus 0 <= u/z' < 1, so that the integer +t = z - u/z' must = [z], the greatest integer <= z. + +Since z generally varies with x and y, it should not +output unless [z] is constant for the range of possible +x and y. We can easily compute the range of z given the ranges +of x and y if we represent each range by the endpoints of +an interval (in either order), along with a bit indicating +Inside or Outside. Thus if z is in standard form, we can say +that z will always be (Inside 1 infinity) (or +(Outside -infinity 1)) after the first term. If z were to +always output its nearest integer instead of its greatest, then +none of the terms after the first would be 1, although they would +probably vary in sign. In this case, z would be (Outside -2 2). + +Now hold y fixed and examine the behavior of z with x. +If x is (Inside a b) then z is (Inside z(a) z(b)) unless +the denominator of z changes sign between a and b (i.e. z has +its pole in this interval), whereupon z is (Outside z(a) z(b)). +Symmetrically, when x is (Outside a b) then z is +(Outside z(a) z(b)) unless the signs of the denominators of +z(a) and z(b) differ, whereupon z is (Inside z(a) z(b)). +This argument still holds with x and y interchanged. + +Now suppose that with y fixed at one of its endpoints, x +constrains z (Inside 1 2), and at y's other extreme, z(x) is +(Outside 0 3). Suppose further that at the two extremes of x, +z(y) is (Inside 1 3) and (Outside 0 2). Then z(x,y) is +(Outside 0 1), the union of the four ranges. (Outside 0 2) is +the widest, indicating that z will probably get more information +from a term of y than a term of x. (Topology hackers should +recognize this Inside-Outside nonsense as ordinary intervals in +toroidal space. The clue is that both plus and minus infinity +are denoted by the empty continued fraction.) + +Due to the basically monotonic +behavior of z, we can guarantee that the actual range of +z will be the union of these four ranges, and that this +range will be Inside or Outside some interval. If it is +(Inside z1 z2) and [z1] = [z2], z can output the term +t = [z1]. Otherwise, z must input a term from x or y, whichever +was associated with the widest of the four ranges of z. +(Outside narrowness) is wider than (Outside wideness) is wider +than (Inside wideness) is wider than (Inside narrowness). + +Evaluating z on these endpoints may be facilitated by +keeping estimates for the integer variables +in floating point. + Even if z doesn't produce a term, narrowing the range of +possible z will still help in computing the range of a function +of z, especially if z gets stuck trying to output the last +term of a rational number resulting from irrational x and y. +(There is no way to guarantee that x or y won't eventually +deviate, whereupon z would egest a gigantic term.) + +z can produce its value as decimal digits by multiplying +by 10 instead of reciprocating, after outputting t = [z]: + + z'(x,y) = (10(a-te) 10(b-tf) 10(c-tg) 10(d-th))/(e f g h). + +Strange to say, it is not serious if z for some reason outputs +the terms 7 5 1 when it should have produced 6 9. As soon as +permitted, it will simply recant with 0 -1 -5 and continue with +the correction -1 9. The sequence 7 5 1 0 -1 -5 -1 9 is +equivalent to 6 9 because b 0 c is the same as b+c. In order to +undo these computations, z violates the condition (Outside -1 1) +when it is 0 -1 -5 ... . This condition is obeyed by nearly all +convergent continued fractions after their first term, and its +violation will very probably cause further retractions among the +functions dependent upon z. + +This computation reversal trick is also handy for mechanizing and +denoting imprecise quantities. Instead of 2.997930 +or- .000003, +we have 2 1 481 0 2, meaning between 2 1 481 and 2 1 483. +Similarly, 137 26 0 1 replaces 137.0373 +or- .0006. + +Successive approximations methods benefit considerably from +not requesting terms until needed. Consider Newton's method +for algebraic roots. We expect successive approximations +to have about twice as many correct terms each time. +Since the production of these terms cannot be aided by +reading incorrect terms, the additional correct terms must +be produced before the bad ones of the previous approximation +are used. But this means that there is no need to read in +the bad ones at all. By feeding back the output terms in +place of the approximation, we get the correct answer directly! +(69% of the credit for this goes to Schroeppel.) + +The basic eight variable form exemplified above by z(x,y) +is not the only form preserved by continued fraction +term transactions. We need only four variables and a +single interval check to compute +z(x) = (ax+b)/(cx+d), the homographic function of one argument. +On the other hand, z(w,x,y) (linear in all three arguments) +requires sixteen variables and a twelve way interval check. +Each of these forms can be solved for x in terms of z etc. +to get a function of the same form. This is not true of + z(x) = (ax^2 + bx + c)/(dx^2 + ex + f), +for example, even though this form is also preserved. +This form is not guaranteed monotone, thus theoretically +invalidating the interval check algorithm, but it hardly ever +errs. Even if it did, it would quickly correct itself anyway. +This form is not only more economical than z(x,x), it is +essential for the success of the Newton's method feedback trick, +which must know when two variables are really the same one. + By choosing the eight coefficients a through h properly, it should +be possible to rewrite arithmetic expressions as compositions of +considerably fewer of these forms than one for each +, -, *, and +/. The reader is invited to investigate the problem of trying +to find minimal representations. Depending on the metric for +minimality, the question can be complicated by allowing higher +powers of x and y. If the highest powers of x, y, z, ... in an +invariant form are i, j, k, ..., then the number of integer +variables required for the coefficients (mostly because of all +of the cross terms) is 2(i+1)(j+1)(k+1)... . + +It is awkward in this system to evaluate transcendental +functions of irrational arguments. The problem is that you may +need any number of continued fraction (or series) terms which, +instead of being numbers, are symbolic functions of x, some +infinite continued fraction. My suggestion is to represent each +symbolic term of the function by a subroutine which is a +function of x and the next term, with this next term really a +dummy until actually called upon for output, whereupon it +replaces itself with a full fledged term subroutine which in +turn refers to x and a new dummy. + +Sad to say, the integer variables in these algorithms do not +usually shrink on outputs as much as they grow on inputs. +Fortunately, the operations for input and output only require +(besides addition) multiplication by terms which are +almost invariably small. (I have not seen a term exceed +20776 except in specially constructed numbers.) It is +fairly safe, then, to declare any function which +has gotten (Outside -2^35 2^35) to be infinite, thus +terminating its continued fraction. Better still, note that the +term 20776 is equivalent to the terms 20000 0 700 0 70 0 6, +i.e., a very large term can be transmitted piecewise. Although +this is just thinly disguised multiprecision multiplication, +that first piece of the term will probably satisfy its recipient +for quite some time. + +In some special cases, the integer variables will become +periodic rather than large, especially when all +but one of the arguments to a function have terminated. +Then, we have the form z(x) = (ax+b)/(cx+d), known as +a homographic function. If ad-bc is +or- 1, then a, b, c, d +will eventually become 1, 0, 0, 1, whereupon z will +output the terms of x unmodified. Periodicity will also +occur when x is a Hurwitz number, i.e., when the terms +of x are the values of one or more polynomials evaluated on +consecutive integers and then interleaved. Coth 1/69, +SQRT 105, and e are Hurwitz numbers whose polynomials +are linear or constant. Hurwitzness is preserved by +homographic functions. If one can show that pi is not +a Hurwitz number, one confirms the long standing conjectures +that e*pi, e+pi, e/pi, etc. are all irrational. + If z, x, and y are all regular, then it generally +won't be possible to reduce z by finding a GCD +of a through h which is greater than one. However, it has been +determined empirically that much reduction is often possible +in other cases. This reduction is almost always by a +divisor of an input or output term numerator (or 10 if output is +decimal digits) and can be facilitated by keeping certain of the +integer variables around modulo these quantities. + +ITEM (Gosper): +Problem: Given an interval, find in it the rational number +with smallest numerator and denominator. +Solution: Express the endpoints as continued fractions. +Find the first term where they differ and add 1 to the +lesser term, unless it's last. +Discard the terms to the right. What's left is the continued +fraction for the "smallest" rational in the interval. +(If one fraction terminates but matches the other as far as +it goes, append an infinity and proceed as above.) + + ********************************************* +GROUP THEORY +********************************************* + +ITEM (Schroeppel): +As opposed to the usual formulation of a group, +where you are given +1 there exists an I such that A * I = I * A = A, and +2 for all A, B and C, (A * B) * C = A * (B * C), and +3 for each A there exists an ABAR such that + A * ABAR = ABAR * A = I, and +4 sometimes you are given that I and ABAR are unique. +If instead you are given A * I = A and A * ABAR = I, then +the above rules can be derived. But if you are given A * I = A +and ABAR * A = I, then something very much like a group, but not +necessarily a group, results. For example, every element is +duplicated. + +ITEM (Gosper): +The Hamiltonian paths through the N! permutations of N objects +using only SWAP (swap any specific pair) and ROTATE (1 position) +are as follows: +N PATHS + DISTINCT REVERSES +2 2 + 0, namely: S, R +3 2 + 1, namely: SRRSR, RRSRR +4 3 + 3, namely: + SRR RSR SRR RSR RRS RSR RSR RR + RSR SRR RSR RRS RSR RRS RSR RR + SRR RSR RRS RRS RSR RRS RRR SR +PROBLEM: A questionable program said there are none for N = 5; +is this so? + +ITEM (Schroeppel): +Any permutation on 72 bits can be coded with a routine +containing only the PDP-6/10 instructions "ROT" and "ROTC". + +********************************************* +SET THEORY +********************************************* + +ITEM (Komolgoroff, maybe?): +Given a set of real numbers, how many sets can you +get using only closure and complement? Answer: 14. + ************************************** +QUATERNIONS +************************************** + +ITEM (Salamin): + A quaternion is a 4-tuple which can be regarded +as a scalar plus a vector. Quaternions add linearly +and multiply (non-commutatively) by + (S1+V1)(S2+V2) = S1 S2 - V1.V2 + S1 V2 + V1 S2 + V1 X V2 +where +S=scalar part, V=vector part, .=dot product, X=cross product. +If Q = S+V = (Q0,Q1,Q2,Q3), then S = Q0, V = (Q1,Q2,Q3). +Define conjugation by (S+V)* = S-V. The (absolute value)^2 +of a quaternion is Q0^2 + Q1^2 + Q2^2 + Q3^2 = Q Q* = Q* Q. + + The non-zero quaternions form a group under multiplication +with (1,0,0,0) = 1 as identity and 1/Q = Q*/(Q* Q). The unit +quaternions, which lie on a 3-sphere embedded in 4-space, form a +subgroup. The mapping F(Q) = PQ (P a unit quaternion) is a rigid +rotation in 4-space. This can be verified by expressing PQ as a +4x4 matrix times the 4-vector Q, and then noting that the matrix +is orthogonal. F(Q) restricted to the unit quaternions is a +rigid rotation of the 3-sphere, and because this mapping is a +group translation, it has no fixed point. + + We can define a dot product of quaternions as the dot +product of 4-vectors. Then Q1.Q2 = 0 iff Q1 is perpendicular to +Q2. Let N be a unit vector. To each unit quaternion Q = S+V, +attach the quaternion NQ = -N.V + N S + N X V. Then it is seen +that (NQ).(NQ) = N.N = 1 and (NQ).Q = 0. Geometrically this +means that NQ is a continuous unit 4-vector field tangent to the +3-sphere. No such tangent vector field exists for the ordinary +2-sphere. Clearly the 1-sphere has such a vector field. +PROBLEM: For which N-spheres does a +continuous unit tangent vector field exist? + Let W be a vector (quaternion with zero scalar part) and +Q = S+V. Then Q W Q* = (S^2 + V.V)W + 2 S V X W + 2 V(V.W). +Let N be a unit vector and Q the unit quaternion +Q = +-(cos(T/2) + N sin(T/2)). Then +Q W Q* = (cos T)W + (sin T)(N X W) + (1-cos T)N(N.W), which is W +rotated thru angle T about N. If Q thus induces rotation R, then +Q1 Q2 induces rotation R1 R2. So the projective 3-sphere +(+Q and -Q identified) is isomorphic to the rotation group +(3x3 orthogonal matrices). Projectiveness is unavoidable since a +2 pi rotation about any axis changes Q = 1 continuously into +Q = -1. + + Let U be a neighborhood of the identity in the rotation +group (ordinary 3 dimensional rotations) and U1 the corresponding +set of unit quaternions in the neighborhood of 1. If a rotation +R carries U into U', then a quaternion corresponding to R carries +U1 into U1'. But quaternion multiplication is a rigid rotation +of the 3-sphere, so U1 and U1' have equal volume. This shows +that in the quaternion representation of the rotation group, the +Haar measure is the Lebesgue measure on the 3-sphere. + + Every rotation is a rotation by some angle T about some +axis. If rotations are chosen "uniformly", what is the +probability distribution of T? By the above, we choose points +uniformly on the 3-sphere (or hemisphere since it is really +projective). Going into polar coordinates, one finds +P(T) = (2/pi) (sin T/2)^2, 0 < T < pi. +In particular, the expected value of T is pi/2+2/pi. + + Quaternions form a convenient 4-parameter representation +of rotations, since composition of rotations is done by +quaternion multiplication. In contrast, 3-parameter +representations like Euler angles or (roll, pitch, yaw) require +trigonometry for composition, and orthogonal matrices are +9-parameter. In space guidance systems under development at +D-lab, the attitude of the spacecraft is stored in the guidance +computer as a quaternion. + ********************************************* +POLYOMINOS, ETC. +********************************************* + +ITEM: +See the PROPOSED COMPUTER PROGRAMS section +for counts of polyominos of orders < 19. + +ITEM (Schroeppel): +Tessellating the plane with polyominos: +Through all hexominos, the plane can be tessellated with +each piece (without even flipping any over). All but the +four heptominos below can tessellate the plane, again without +being flipped over. Thus, flipping does not buy you anything +through order 7. (There are 108 heptominos). + H H HHH H H + HHHHH H H HHHH HHHH + HH H H + H H + +ITEM (Schroeppel): +PROBLEM: What rectangles are coverable +by various polyominos? For example, + + XX can cover rectangles which are 3N by M, + X except if N = 1, then M must be even. + + YYYY can be shown by coloring to cover only rectangles + having at least one side divisible by four. + +ITEM (Schroeppel): +PROBLEM: Find a necessary and sufficient condition for +an arbitrary shape in the plane to be domino coverable. + ITEM (Beeler): +"Iamonds" are made of equilateral triangles, like diamonds. +"(Poly-)ominos" are made of squares, like dominos. +"Hexafrobs" are made of hexagons. +"Soma-like" pieces are made of cubes. +See also "Polyiamonds," Math. Games, Sci. Am., December 1964. +Left and right 3-dimensional forms are counted as distinct. +ORDER IAMONDS OMINOS HEXA'S SOMA-LIKE + 1 1 1 1 1 + 2 1 1 1 1 + 3 1 2 3 2 + 4 3 5 7 8 + 5 4 12 22 29 + 6 12 35 + 7 24 + 8 66 + 9 160 +10 448 +Polyominos of order 1, 2 and 3 cannot form a rectangle. Orders +4 and 6 can be shown to form no rectangles by a checkerboard +coloring. Order 5 has several boards and its solutions are +documented (Communications of the ACM, October 1965): + BOARD DISTINCT SOLUTIONS + 3 X 20 2 + 4 X 15 368 + 5 X 12 1010 + 6 x 10 2339 (verified) + two 5 X 6 -- 2 + 8 X 8 with 2 X 2 hole in center -- 65 +CONJECTURE (Schroeppel): If the ominos of a given order +form rectangles of different shapes, the rectangle which +is more nearly square will have more solutions. +Order-4 hexafrob boards and solution counts: + side 7 triangle -- no solutions + parallelogram, base 7, side 4 -- 9 distinct solutions +e.g., A A A A B C C + D E B B C F C + D E E B F G G + D D E F F G G +Order-6 iamond boards and solution counts (see illustration): + side 9 triangle with inverted side 3 triangle + in center removed -- no solutions + trapezoid, side 6, bases 3 and 3+6 -- no solutions + two triangles of side 6 -- no solutions + trapezoid, side 4, bases 7 and 7+4 + -- 76 distinct solutions + parallelogram, base 6, side 6 -- 156 distinct solutions + parallelogram, base 4, side 9 -- 37 distinct solutions + parallelogram, base 3, side 12 -- no solutions + triangle of side 9 with triangles of side 1, 2 and 2 + removed from its corners (a commercial puzzle) + -- 5885 distinct solutions + With Soma-like pieces, orders 1, 2 and 3 do not have interesting +boxes. Order 4 has 1390 distinct solutions for a 2 X 4 X 4 box. +1124 of these have the four-in-a-row on an edge; the remaining +266 have that piece internal. 320 solutions are due to +variations of ten distinct solutions decomposable into +two 2 X 2 X 4 boxes. A Soma-like 2 X 4 X 4 solution: + AAAA BBHH + BCCC BHHC + DDDE FGGE + FDGE FFGE +The commercial Soma has 240 distinct solutions; the booklet +which comes with it says this was found years ago on a 7094. +Verified by both Beeler and Clements. + ********************************************* +TOPOLOGY +********************************************* + +ITEM: +Although not new (cf Coxeter, Introduction to Geometry, +1st ed. p393), the following coloring number +(chromatic number) may be useful to have around: + N = [[(7 + SQRT (1 + 48 * H))/2]] +where N is the number of colors required to color any map on an +object which has H holes (note: proof not valid for H = 0). +For example: +A donut (holes = 1) requires 7 colors to color maps on it. +A 17-hole frob requires 17 colors. +An 18-hole frob requires 18 colors. + +ITEM (Schroeppel): +A most regular 7-coloring of the torus can be +made by tiling the plane with the following +repeating pattern of hexagons of 7 colors: + + A A C C E E +A A A C C C E E E + A A F F C C A A E E + F F F A A A + B B F F D D A A F F +B B B D D D F F F + B B G G D D B B F F + G G G B B B + C C G G E E B B G G +C C C E E E G G G + C C A A E E C C G G + A A A C C C + D D A A F F C C A A +D D D F F F A A A + D D B B F F D D A A + B B B D D D + E E B B G G D D B B +E E E G G G B B B + E E C C G G E E B B + C C C E E E + C C E E +Draw an area 7 unit cell parallelogram by connecting, say, the +center B's in each of the four B B + B B B + B B . Finally, join the +opposite sides of the parallelogram to form a torus in the +usual (Spacewar) fashion. QUESTION (Gosper): is there a toroidal +heptahedron corresponding to this? + ITEM (Gosper): +A spacefilling curve is a continuous map T --> X(T),Y(T), +usually from the unit interval onto the unit square, +often presented as the limit of a sequence of curves made by +iteratively quadrisecting the unit square. Each member of the +sequence is then 4 copies of its predecessor, connected in the +shape of an inverted V, with the first member being a V which +connects 0,0 to 1,0. The limiting map, X(T) and Y(T), can be +computed instead by a simple, finite-state machine having 4 +inputs (digits of T base 4), 4 outputs (one bit of X and one bit +of Y), and 4 states (2 bits) of memory (the number modulo 2 of +0's and 3's seen in T). + Let T, X, and Y be written in binary as: + +T=.A B A B A B ... X=.X X X X X X ... Y=.Y Y Y Y Y Y ... + 1 1 2 2 3 3 1 2 3 4 5 6 1 2 3 4 5 6 + +ALGORITHM S: + + C _ 0 ;# of 0's mod 4 + 0 + + C _ 0 ;# of 3's mod 4 + 1 + +S1: X _ A XOR C ;Ith bit of X + I I NOT B + I + Y _ X XOR B ;Ith bit of Y + I I I + + C _ C XOR (NOT A AND NOT B ) ;count 00's + 0 0 I I + + C _ C XOR (A AND B ) ;count 11's + 1 1 I I + + GO S1 + OLD NEW +C C A B X Y C C + 0 1 I I I I 0 1 + +0 0 0 0 0 0 1 0 +0 0 0 1 0 1 0 0 +0 0 1 0 1 1 0 0 +0 0 1 1 1 0 0 1 +0 1 0 0 1 1 1 1 +0 1 0 1 0 1 0 1 This is the complete +0 1 1 0 0 0 0 1 state transition table. +0 1 1 1 1 0 0 0 +1 0 0 0 0 0 0 0 +1 0 0 1 1 0 1 0 +1 0 1 0 1 1 1 0 +1 0 1 1 0 1 1 1 +1 1 0 0 1 1 0 1 +1 1 0 1 1 0 1 1 +1 1 1 0 0 0 1 1 +1 1 1 1 0 1 1 0 + +To carry out either the forward or reverse map, label a set of +columns as in the table above. Fill in whichever you know of AB +or XY, with consecutive rows corresponding to consecutive I's. +Put 0 0 in the top position of the OLD CC column. Exactly one +row of the above table will match the row you have written so +far. Fill in the rest of the row. Copy the NEW CC entry to the +OLD CC column in the next row. Again, only one row of the state +table will match, and so forth. For example, the map +5/6 --> (1/2,1/2) (really .11010101... --> (.1000... ,.0111...)): + + OLD NEW +C C A B X Y C C + 0 1 I I I I 0 1 + +0 0 1 1 1 0 0 1 +0 1 0 1 0 1 0 1 +0 1 0 1 0 1 0 1 +0 1 0 1 0 1 0 1 +. . . . . . . +. . . . . . . + = 5/6 1/2 1/2 + +We note that since this is a one-to-one map on bit strings, it is +not a one-to-one map on real numbers. For instance, there are 2 +ways to write 1/2, .1000... and .0111..., and thus 4 ways to +write (1/2,1/2), giving 3 distinct inverses, 1/6, 1/2, and 5/6. +Since the algorithm is finite state, X and Y are rational iff T +is, e.g., 898/4369 --> (1/5,1/3). The parity number, (see SERIES +section) and 1-(parity number) are the only reals satisfying +X(T)=T, Y(T)=1. This is related to the fact that they have no +0's and 3's base 4, and along with 0, 1/2, and 1=.111..., are the +only numbers preserved by the deletion of their even numbered bit +positions. + ********************************************* +SERIES +********************************************* + +ITEM (Schroeppel & Gosper): +The sum from N = 0 of [N!N!/(2N)!] = 4/3 + 2*pi/9*SQRT 3. +PROBLEM: Evaluate in closed form the sum from N = 0 of +[N!N!N!/(3N)!]. This sum has been shown equal to the following: +the integral from 0 to 1 of (P + Q arccos (R))dT +where + 2 (8 + 7 T^2 - 7 T^3) +P = --------------------- + (4 - T^2 + T^3)^2 +and + 4 (T - T^2) (5 + T^2 - T^3) +Q = ------------------------------------------------ + (4 - T^2 + T^3)^2 SQRT ((4 - T^2 + T^3) (1 - T)) +and + T^2 - T^3 +R = 1 - --------- . + 2 + +ITEM (Euler): +The series accelerating transformation +(Abramowitz & Stegun, sec. 3.6.27) + K K K+1 +A0-A1+A2-... = SUM (-) (DELTA A0)/2 + + K K-M +(where (DELTA A0) = SUM M=0 to K (-) (BINOMIAL K M) AM = Kth +forward difference on A0) when applied to + N-1 2 +pi/4 = 1 - 1/3 + 1/5 - ... gives pi/4 = SUM 2 N! /(2N+1)! . + +Applied to the formula for gamma in Amer. Math. Monthly + T +(vol. 76, #3, Mar69 p273) = SUM (-) [LOG2 T]/T +(brackets mean integer part of) we get + -(K+1) I +SUM K=1 to infinity 2 SUM J=0 to K-1 1/(BINOMIAL 2 +J J) +(Gosper) which converges fast enough for a few hundred digits. +The array of reciprocals of the terms follows, with powers of 2 +factored out to the left from all members of each row. +4 1 +8 1 3 +16 1 5 6 +32 1 9 15 10 +64 1 17 45 35 15 +128 1 33 153 165 70 21 +256 1 65 561 969 495 126 28 + + N+1 +The next to left diagonal is 2 ; the perpendicular one 3rd +from the right is 1, *9/1= 9, *10/2= 45, *11/3= 165, *12/4= 495. + ITEM (Henry Cohen): +Gamma = - LN X + X - X^2/2*2! + X^3/3*3! - X^4/4*4! ... + ERROR + -X +Where ERROR is of the order of (e )/X. + +ITEM (Schroeppel): -Y + Differentiate Ye = X to get Y+YXY'-XY' = 0. +Substitute for Y a power series in X with coefficients to be determined. +One observes the curious identity: + J-1 N-J N +SUM J=1 to N of (BINOMIAL N J)J (N-J) = N (0^0=1) + N-1 N +and thus Y(X) = SUM N=1 N X /N! + +ITEM (Schroeppel): +PROBLEM: Can someone square some series for pi to give +the series pi^2/6 = 1/1^2 + 1/2^2 + ... = SUM 1/N^2 ? + +ITEM (Schroeppel): +Consider SUM 1/N^2 = + SUM 1/(N-1/2)-1/(N+1/2) + SUM 1/N^2-1/(N^2-1/4) + = 2 - SUM 1/((4N^2-1)*N^2). +Take the last sum and re-apply this transformation. +This may be a winner for computing the original sum. +For example, the next iteration gives + 31/18 - SUM 9/(N^2)(4N^2-1)(25N^4+5N^2+9) + where the denominator also = + (N^2)(2N+1)(2N-1)(5N^2+5N+3)(5N^2-5N+3) + +ITEM (Polya): +CONJECTURE: If a function has a power series with integer +coefficients and radius of convergence 1, then either the +function is rational or the unit circle is a natural +boundary. Reference: Polya, Mathematics and Plausible +Reasoning, vol. 2, page 46. + +ITEM (Gosper): +Consider the triangular array: + 1 + 1 1 + 1 4 1 + 1 11 11 1 + 1 26 66 26 1 + 1 57 302 302 57 1 +This bears an interesting relationship to Pascal's triangle. +The 302 in the 4th southeast diagonal and the 3rd southwest one += 4*26 + 3*66. Note that rows then sum to factorials rather +than powers of 2. If the nth row of the triangle is dotted with +any n consecutive elements of (either) n+1st diagonal of Pascal's +triangle, we get the nth Bernoulli polynomial: for n = 5, +1(6,i) + 26(6,i+1) + 66(6,i+2) + 26(6,i+3) + 1(6,i+4) = +sum of 5th powers of 1 thru i+5, where (j,i) = Binomial (j+i j). + ITEM (Schroeppel, Gosper): inf -N + The "parity number" = 1/2 SUM (parity of N)*2 + N=0 +where the parity of N is the sum of the bits of N mod 2. +The parity number's value is .4124540336401075977..., or, +for hexadecimal freaks, .6996966996696996... . It can be +written (base 2) in stages by taking the previous stage, +complementing, and appending to the previous stage: + .0 + .01 + .0110 + .01101001 + .0110100110010110 + .01101001100101101001... radix 2 + + N N +i.e., stage 0 = 0 -2 -2 + stage N+1 = stage N + (1-2 -stage N)/2 . + +If NUM 0 = 0, DEN 0 = 2 + NUM N+1 = ((NUM N)+1)*((DEN N)-1) + N+1 + 2 2 + DEN N+1 = (DEN N) = 2 + N N + NUM N+1 -2 -2 +then ------- = stage N+1 = (stage N + 2 )*(1 - 2 ) . + DEN N+1 + +Or, faster, by substituting in the string at any stage: + the string itself for zeros, and + the complement of the string for ones. +It is claimed (perhaps proven by Thue?) +that the parity number is transcendental. + +Its regular continued fraction begins: 0 2 2 2 1 4 3 5 2 1 4 2 1 +5 44 1 4 1 2 4 1 1 1 5 14 1 50 15 5 1 1 1 4 2 1 4 1 43 1 4 1 2 1 +3 16 1 2 1 2 1 50 1 2 424 1 2 5 2 1 1 1 5 5 2 22 5 1 1 1 1274 3 5 +2 1 1 1 4 1 1 15 154 7 2 1 2 2 1 2 1 1 50 1 4 1 2 867374 1 1 1 5 +5 1 1 6 1 2 7 2 1650 23 3 1 1 1 2 5 3 84 1 1 1 1284 ... and seems +to continue with sporadic large terms in suspicious patterns. +A non-regular fraction is +1/(3 -1/(2 -1/(4 -3/(16 -15/(256 -255/(65536 -65535/ + N N + 2 2 + (...2 -(2 -1)/(... . +This fraction converges much more rapidly than the regular one, +its Nth approximant being (1+NUM N)/(1+DEN N), which is, in fact, + N +an approximant of the regular fraction, roughly the 2 th. + In addition, 4*(parity number) = + 2-(1/2)*(3/4)*(15/16)*(255/256)*(65535/65536)*... +This gives still another non-regular fraction per the product +conversion item in the CONTINUED FRACTION section. + +For another property of the parity number, see the +spacefilling curve item in the TOPOLOGY section. + +ITEM (Schroeppel, Gosper, Salamin): +Consider the image of the circle ABS z = 1 under the function + n + 2 + z +f(z) = SUM --- . This is physically analogous to a series of + n + 2 +clock hands placed end to end. The first hand rotates around the +center (0,0) at some rate. The next hand is half as long and +rotates around the end of the first hand at twice this rate. +The third hand rotates around the end of the second at four times +this rate; etc. It would seem that the end of the "last" hand +(really there are infinitely many) would sweep through space very +fast, tracing out an (infinitely) long curve in the time the +first hand rotates once. The hands shrink, however, because of + n +the 2 in the denominator. Thus it is unclear whether the speed +of the "last" hand is really infinite; or, whether the curve's +arc length is really infinite. + n! + z +Also, it is a visually interesting curve, as are f(z) = SUM --- , + FIB(n) n! + z +and f(z) = SUM ------- . Gosper has programmed the one mentioned + FIB(n) +first, which makes an intriguing display pattern. See following +illustrations. If you write a program to display this, be sure +to allow the following variations: + _ +(1) z and z on alternate terms (alternate hands rotate in + opposite directions), +(2) negation of alternate terms (alternate hands initially + point in opposite directions), and +(3) how many terms are used in the computation, +since these cause fascinating variations in the resulting curve. + ********************************************* +FLOWS AND ITERATED FUNCTIONS +********************************************* + +ITEM (Schroeppel): +An analytic flow for Newton's method square root: +Define F(X) by (X^2+K)/2X; then + N N + 2 2 + (X+SQRT K) + (X-SQRT K) +F(F(F(...(X)))) = SQRT K --------------------------- + [N times] N N + 2 2 + (X+SQRT K) - (X-SQRT K) + + N +which = SQRT K (coth 2 (arccoth X/SQRT K)) + +ITEM (Schroeppel): +P and Q are polynomials in X; when does P(Q(X)) = Q(P(X)) ? +(That is, P composed with Q = Q composed with P.) +Known solutions are: +1 Various linear things. +2 X to different powers, sometimes multiplied by roots of 1. +3 P and Q are each another polynomial R composed + with itself different numbers of times. +4 Solutions arising out of the flow of X^2-2, as follows: + suppose X = Y + 1/Y + N -N + then Y + Y can be written as a polynomial in X + for example, + P = the expression for squares = X^2-2 (N = 2) + and Q = the expression for cubes = X^3-3X (N = 3) +5 Replace X by Y-A, then add A to the original constants + in both P and Q. For example, P = X^2 and Q = X^3, + then P = 1+(Y-1)^2 = Y^2-2Y+2 and Q = 1+(Y-1)^3, + then P(Q) = 1+(Y-1)^6 = Q(P). + Similarly, replacing X with AY+B works. +6 There are no more through degrees 3 and 4 (checked with + Mathlab); but are there any more at all? + ITEM (Schroeppel): +PROBLEM: Given F(X) as a power series in X +with constant term = 0, write the flow power series. +FLOW sub ZERO = X +FLOW sub ONE = F(X) +FLOW sub TWO = F(F(X)) +etc. +NOTE (Gosper): If we remove the restriction that F has a power +series, the functions that satisfy an equation of the form +F(F(X)) = sin X can be put into one-to-one correspondence with +the set of all functions. + +ITEM (Salamin): P + N N P +If F(X)=X , the P-th flow is X , which has a branch point if N +is non-integer. Under the hypotheses of the previous problem, +it is possible to find the power series coefficients for +P rational, but there is no guarantee the series will converge. +PROBLEM: Is the flow interpolation unique? If it is not, what +extra conditions are necessary to make it unique for natural + N +cases like X ? + ITEM (Schroeppel): +Taking any two numbers A and B, finding their arithmetic mean and +their geometric mean, and using these means as a new A and B, +this process, when repeated, will approach a limit which can be +expressed in terms of elliptic integrals. (See PI section) + +ITEM (Gosper): LOOP DETECTOR +If a function F maps a finite set into itself, then its flow must +always be cyclic. If F is one step of a pseudorandom number +generator, or the CDR operation on a self referent list, or any +function where it is easy to supply former values as arguments, +then there are easy ways to detect looping of the flow (Knuth, +The Art of Computer Programming, volume 2, Seminumerical +Algorithms, sec. 3.1, prob. 7, page 7). If, however, the process +of iterated application of the function is inexorable, +(i.e., there is no easy way to switch arguments to the function), +then the following algorithm will detect repetition before +the third occurrence of any value. + +Set aside a table TAB(J), 0 <= J <= LOG2 (largest possible period). +Let C = the number of times F has been applied, initially 0. +Compare each new value of F for equality with those table entries +which contain old values of F. These will be the first S +entries, where S is the number of times C can be right shifted +before becoming 0. No match means F hasn't been looping very +long, so increment C and store this latest value of F into +TAB(J), where J is the number of trailing zero bits in the binary +of C. (The first 16 values of J are: 0, 1, 0, 2, 0, 1, 0, 3, 0, +1, 0, 2, 0, 1, 0, 4, ...; Eric Jensen calls this the RULER +function.) A match with entry E means the loop length is 1 more + E+1 +than the low E+2 bits of C - 2 . + +ITEM (Schroeppel, Gosper, Henneman & Banks) (from Dana Scott?): +The "3N+1 problem" is iteratively replacing N by N/2 if N is even +or by 3N+1 if N is odd. Known loops for N to fall into are: +1 the zero loop, 0 => 0 +2 a positive loop, 4 => 2 => 1 => 4 +3 three negative loops + (equivalent to the 3N-1 problem with positive N) + -2 => -1 => -2 + -5 => -7 => -10 => -5 + -17 => -25 => -37 => -55 => -82 => -41 => + -61 => -91 => -136 => -68 => -34 => -17 +In the range -10^8 < N < 6 * 10^7, all N fall into the above loops. +Are there any other loops? Does N ever diverge to infinity? + +ITEM (Schroeppel, Gosper): +Let N be iteratively replaced by (FLATSIZE (LONGHAND N)), +the number of letters in N written longhand +(e.g., 69 => SIXTY NINE => 9 (10 counting blanks)). +The process invariably loops at 4 = FOUR. + ITEM (Gosper): +The "C" Curve +A brilliant archeologist is photographing a strange drawing +on the wall of a cave. He holds the camera upright for some +shots, moves it, and turns it 90 degrees for the rest. When he sees +his prints he is amazed to find one of them apparently taken with +the camera turned 45 degrees. After a moment's reflection, he +correctly concludes that it is merely a double exposure. +What was the drawing? + +Answer: It is a cousin to both the dragon and snowflake curves +(and arose as a bug in a spacefilling curve). It can be +constructed as follows. Start with a line segment. +Replace it with the two legs of the isosceles right triangle +of which it is hypotenuse. Repeat this for the two new segments, +always bulging outward in the same direction. We now have four +segments forming half a square, with the middle two segments +collinear. Replacing these four segments with eight and then +sixteen, we find the middle two segments superimposed. +As the process continues, the curve crosses itself more and more +often, eventually taking on the shape of a wildly curly letter C +which forms the envelope of a myriad of epicyclic octagons. + +A faster way to approach the same limiting curve is to substitute +the curve itself for each of its 2^2^n segments, starting with a +90 degree "<". + +Yet another way to construct it is to iteratively connect +opposite ends of two copies at a 90 degree angle. (The +archeologist did this with his double exposure.) If we reduce +the scale by SQRT 2 each time, the distance between the endpoints +stays the same. If the initial line segment is red and there is +some other blue shape elsewhere in the picture, the iteration +will simultaneously proliferate and shrink the blue shapes, until +they are all piled up along the red "C". Thus, no matter what +you start with, you eventually get something that looks like the +"C" curve. + +There are other pictures besides the C curve which are preserved +by this process, but they are of infinite size. You can get +them by starting with anything and running the iteration +backwards as well as forwards, superimposing all the results. A +backward step consists of rotating the two copies in directions +opposite those in the forward step and stretching by SQRT 2 instead +of shrinking. David Silver has sketched an arrangement of +mirrors which might do this to a real scene. + + + **************************************** +PI +**************************************** + +ITEM: GAUSSIAN INTEGERS (For use by next item.) +Reference: Hardy and Wright, Theory of Numbers. The Gaussian +integers are x+iy where x and y are integers. Unique +factorization holds, except for powers of i, and the Gaussian +primes are (1) a+bi if a^2+b^2 is prime and (2) integer primes +that = 3 mod 4. If N(x+iy) = x^2+y^2, then N(uv) = N(u) N(v). +If p is prime and not= 3 mod 4, then p = a^2+b^2 has exactly one +solution. If n = 3 mod 4, then n = a^2+b^2 has no solution. +To factor x+iy into Gaussian primes, first factor N(x+iy). + (A) If 2 divides N(x+iy), then 1+i and 1-i divide x+iy. + Either factor may be used since i(i-1) = i+1. + (B) If p=3 mod 4 divides N(x+iy), then p divides x+iy. + (C) If p=1 mod 4 divides N(x+iy) and p = a^2+b^2, + then a+ib or b+ia = i(a-ib) divides x+iy. + If both do, then p divides x+iy. + +ITEM (Salamin): GENERATION OF ARCTANGENT FORMULAS FOR PI +n1 atan(y1/x1) + n2 atan(y2/x2) + ... + = n1 arg(x1+iy1) + n2 arg(x2+iy2) + ... +If each x+iy is factored and the n's chosen so all prime factors +except 1+i cancel out, the right hand side is a multiple K of +pi/4. Some care is needed because of the multiple valuedness of +arg. Then, if K = 0, we get an arctangent identity, otherwise we +get a pi formula. In the special case of atan(1/x), factorization +of x+i is needed. Then case (B) above can't occur, and in case +(C), a+ib and a-ib can't both divide x+i. +Example: + 8^2+1 = 13 * 5 + 18^2+1 = 13 * 5^2 + 57^2+1 = 13 * 5^3 * 2 +From this we get the factorization + 8+i = (3+2i) (2-i) + 18+i = (3-2i) (2-i)^2 i + 57+i = (3-2i) (2+i)^3 (1-i) +Since we only care about the phase, multiplication +by a positive real number may be ignored below. + a b c + (8+i) (18+i) (57+i) = + + a-b-c -a-2b+3c c b + (3+2i) (2+i) (1-i) i + +We require a-b-c = 0 and -a-2b+3c = 0, which has the minimal +non-trivial solution a = 5, b = 2, c = 3. Then we have + (8+i)^5 (18+i)^2 (57+i)^3 = (1-i)^3 i^2 +Taking the phase of both sides, we get + 5 atan(1/8) + 2 atan(1/18) + 3 atan(1/57) = pi/4. + Pi formulas: + pi/4 = atan(1/2) + atan(1/3) + pi/4 = 2 atan(1/3) + atan(1/7) + pi/4 = 4 atan(1/5) - atan(1/239) + pi/4 = 2 atan(1/4) + atan(1/7) + 2 atan(1/13) + pi/4 = 3 atan(1/4) + atan(1/13) - atan(1/38) + pi/4 = 4 atan(1/5) - atan(1/70) + atan(1/99) + pi/4 = 5 atan(1/8) + 2 atan(1/18) + 3 atan(1/57) + pi/2 = 7 atan(1/4) - 5 atan(1/32) + 3 atan(1/132) - 4 atan(1/378) +This last angle has been measured against the International +Standard Platinum-Iridium Right Angle and certified adequate +for any purpose of the U. S. Government, when used in conjunction +with a conscientiously applied program of oral hygiene and +regular professional care. + pi/4 = 7 atan(1/9) + atan(1/32) - 2 atan(1/132) - 2 atan(1/378) + pi/4 = 7 atan(1/13) + 8 atan(1/32) - 2 atan(1/132) + 5 atan(1/378) + +There are many easily found arctangent identities. Some are: + atan(1/31) = atan(1/57) + atan(1/68) + = atan(1/44) + atan(1/105) + atan(1/50) = atan(1/91) + atan(1/111) + atan(1/239) = atan(1/70) - atan(1/99) + = atan(1/408) + atan(1/577) + atan(1/2441) = atan(1/1164) - atan(1/2225) + = atan(1/4774) + atan(1/4995) + atan(1/32) = atan(1/38) + atan(1/132) - atan(1/378) + = 2 atan(1/73) + atan(1/239) - atan(1/2943) + +Infinite sets of arctangent identities: + atan(1/n) - atan(1/(n+1)) = atan(1/(n^2+n+1)) + Let x =1, y =0, x =x +2y , y =x +y . + 0 0 n n-1 n-1 n n-1 n-1 + + x /y are the continued fraction approximants to SQRT(2). + n n + atan(1/y ) + atan(1/x ) = atan(1/x ) + 2n 2n 2n-1 + + atan(1/y ) - atan(1/x ) = atan(1/x ) + 2n 2n 2n+1 + +ITEM (Gosper): +pi = 28 ARCTAN (3/79) + 20 ARCTAN (29/278) +pi = 48 ARCTAN (3/79) + 20 ARCTAN (1457/22049) +Which isn't too interesting except that it means that +(79+3i)^48 (22049+1457i)^20 is a negative real number. + ITEM (Ramanujan): +4/pi = SUM from N=0 to infinity + + N + (-1) (1123 + 21460 N) (1*3*5*...*(2N-1)) (1*3*5*...*(4N-1)) + ------------------------------------------------------------ + 2N+1 N + (882 ) (32 ) (N!)^3 + +This series gives about 6 decimal places accuracy per term. + +1/(sqrt(8) pi) = SUM from N=0 to infinity + + (1103 + 26390 N) (1*3*5*...*(2N-1)) (1*3*5*...*(4N-1)) + ------------------------------------------------------ + 4N+2 N + (99 ) (32 ) (N!)^3 + +This series gives about 8 decimal places accuracy per term. +For other pi series, see Ramanujan's paper "Modular Equations and +Approximations to Pi" in Quarterly Journal of Pure and Applied +Mathematics, vol. 45, page 350 (1914). For more goodies, see +"Collected Papers of Srinivasa Ramanujan", Cambridge U. Press +(1927). + +ITEM: +Counting the initial 3 as the zeroth, +the 431st denominator in the regular continued +fraction for pi is 20776. (Choong, Daykin & +Rathbone, Math. of Computation 25 (1971) p. 387). +(Gosper) In the first 26491 terms of pi, the only other 5 digit +terms are the 15543rd = 19055 and the 23398th = 19308. (Computed from +35570 terms of the (nonregular) fraction for 4 arctan 1.) + +ITEM: +The fraction part of pi 10^760 begins: .49999998... + +ITEM (Salamin): +Some super-fast convergents to pi if one already +has a super-fast computation of trig functions. + X approx pi: X _ X + sin X, EPSILON _ EPSILON^3/6 + X _ X - tan X, EPSILON _ -EPSILON^3/3 + X approx pi/2: X _ X + cos X, EPSILON _ EPSILON^3/6 + X _ X + cot X, EPSILON _ -EPSILON^3/3 + ITEM (Salamin): +Computation of elliptic integrals, log, and pi. +REFERENCES: +Whittaker & Watson, Modern Analysis, chap. 22 +Abramowitz & Stegun, Handbook of Mathematical Functions, + sect. 17.3, 17.6 + +1. ELLIPTIC INTEGRALS + Define elliptic integrals: + 1 2 2 + K(m) = INTEGRAL 1/SQRT[(1 - t ) (1 - m t )] dt + 0 + K'(m) = K(1 - m) + If A and B are given, and + 0 0 + A = arithmetic mean of A and B + n+1 n n + B = geometric mean of A and B , + n+1 n n +then define + AGM(A ,B ) = lim A = lim B + 0 0 n n +This is called the arithmetic-geometric mean. +Quadratic convergence rate: + A - B = (A - B )^2/8A + n+1 n+1 n n n+1 + + K'(x^2) AGM(1,x) = pi/2 [see A&S]. +This gives a super fast method of computing elliptic integrals. +It is easy to compute AGM(1,x) for x in the complex plane cut from +zero to infinity along the negative real axis. So K'(m) can be +computed for -2 pi < arg(m) < 2 pi, which covers the complex +m-plane twice. Handling the phase when taking square roots will +permit exploration of more of the Riemann surface. + 2. LOGARITHMS + For small m, + K(m) = (pi/2) (1 + m/4 + O(m^2)) + exp[-pi (K'(m)/K(m))] = (m/16) (1 + m/2 + O(m^2)) + +Solve for K'(m) and let m = 16/x^2, + K'(16/x^2) = log x + (4/x^2) (log x - 1) + O(log x/x^4). +For x sufficiently large, + log x = K'(16/x^2) = pi/(2 AGM(1, 4/x)). + +Requiring a given number of bits accuracy in log x is equivalent +to requiring + ABS[(K'(16/x^2) - log x)/log x] < EPSILON. +This becomes + ABS[(4/x^2) (1 - 1/log x)] < ABS(4/x^2) < EPSILON + ABS(x) > 2/SQRT(EPSILON). + +x can be complex. If ABS(x) is not too close to 1, x can be +brought into range by reciprocating or repeated squaring. + +3. PI n + Let x = e , then + -n + pi = 2 n AGM(1, 4 e ). + +Suppose EPSILON = 10 to the minus a billion. +Then the above equation for pi is valid when n > 1.15 billion. + -n +e is calculated by starting with 1/e and squaring k times. + k +Thus n = 2 . 2^30 = 1.07 billion and 2^31 = 2.15 billion, +so k = 30 gives 0.93 billion places accuracy and k = 31 gives +1.86 billion places. + +ITEM (Schroeppel): + n n n +In the above, instead of x = e , use x = 2 and x = e*2 . +Then simultaneous equations can be solved to give both pi and +log 2. This avoids having to square e, but requires two AGM's, +and therefore takes longer. + ********************************************* +PROGRAMMING HACKS +********************************************* + +WARNING: Numbers in this section are octal +(and occasionally binary) unless followed by a decimal point. +105=69.. (And 105.=69 hexadecimal.) + +ITEM (Gosper): +Proving that short programs are neither trivial nor +exhausted yet, there is the following: +0/ TLCA 1,1(1) +1/ see below +2/ ROT 1,9 +3/ JRST 0 +This is a display hack (that is, it makes pretty patterns) with +the low 9 bits = Y and the 9 next higher = X; also, it makes +interesting, related noises with a stereo amplifier hooked to +the X and Y signals. Recommended variations include: + CHANGE: GOOD INITIAL CONTENTS OF 1: + none 377767,,377767; 757777,,757757; etc. + TLC 1,2(1) 373777,,0; 300000,,0 + TLC 1,3(1) -2,,-2; -5,,-1; -6,,-1 + ROT 1,1 7,,7; A0000B,,A0000B + ROTC 1,11 ;Can't use TLCA over data. + AOJA 1,0 + +ITEM (Cohen): +Another simple display program: ("munching squares") +It is thought that this was discovered by +Jackson Wright on the RLE PDP-1 circa 1962. + DATAI 2 + ADDB 1,2 + ROTC 2,-22 + XOR 2,1 + JRST .-4 +2=X, 3=Y. Try things like 1001002 in data switches. This also +does interesting things with operations other than XOR, +and rotations other than -22. (Try IOR; AND; TSC; FADR; FDV(!); +ROT -14, -9, -20, ...) + +ITEM (Schroeppel): +Munching squares is just views of the graph Y = X XOR T +for consecutive values of T = time. + +ITEM (Cohen, Beeler): +A modification to munching squares which +reveals them in frozen states through +opening and closing curtains: insert FADR 2,1 before the XOR. +Try data switches = +4000,,4 1000,,2002 2000,,4 0,,1002 +(Notation: ,,) +Also try the FADR after the XOR, switches = 1001,,1. + ITEM (Minsky): +Here is an elegant way to draw almost circles on +a point-plotting display. CIRCLE ALGORITHM: + NEW X = OLD X - EPSILON * OLD Y + NEW Y = OLD Y + EPSILON * NEW(!) X +This makes a very round ellipse centered at the origin with its +size determined by the initial point. EPSILON determines the +angular velocity of the circulating point, and slightly affects +the eccentricity. If EPSILON is a power of 2, then we don't even +need multiplication, let alone square roots, sines, and cosines! +The "circle" will be perfectly stable because the points soon +become periodic. + +The circle algorithm was invented by mistake when I tried to +save one register in a display hack! Ben Gurley had an amazing +display hack using only about six or seven instructions, and it +was a great wonder. But it was basically line-oriented. +It occurred to me that it would be exciting to have curves, and I +was trying to get a curve display hack with minimal instructions. + +ITEM (Schroeppel): +PROBLEM: Although the reason for the circle algorithm's +stability is unclear, what is the number of distinct sets +of radii? (Note: algorithm is invertible, so all points +have predecessors.) + +ITEM (Gosper): +Separating X from Y in the above recurrence, + X(N+1) = (2-EPS^2)*X(N) - X(N-1) + Y(N+1) = (2-EPS^2)*Y(N) - Y(N-1). +These are just the Chebychev recurrence with cos THETA (the +angular increment) = 1-EPS^2/2. Thus X(N) and Y(N) are expressible +in the form R cos(N THETA + PHI). The PHI's and R for X(N) and Y(N) +can be found from N=0,1. The PHI's will differ by less than pi/2 +so that the curve is not really a circle. The algorithm is +useful nevertheless, because it needs no sine or square root +function, even to get started. + +X(N) and Y(N) are also expressible in closed form in the +algebra of ordered pairs described under linear recurrences, +but they lack the remarkable numerical stability of the +"simultaneous" form of the recurrence. + +ITEM (Salamin): +With exact arithmetic, the circle algorithm is stable iff +ABS(EPSILON)<2. In this case, all points lie on the ellipse + X^2 - EPSILON*X*Y + Y^2 = constant, +where the constant is determined by the initial point. +This ellipse has its major axis at 45 degrees (if EPSILON > 0) +or 135 degrees (if EPSILON < 0) and has eccentricity + SQRT (EPSILON/(1 + EPSILON/2)). + ITEM (Minsky): +To portray a 3-dimensional solid on a 2-dimensional display, +we can use a single circle algorithm to compute orbits for the +corners to follow. The (positive or negative) radius of each +orbit is determined by the distance (forward or backward) from +some origin to that corner. The solid will appear to wobble +rigidly about the origin, instead of simply rotating. + +ITEM (Gosper): +The myth that any given programming language is machine +independent is easily exploded by computing the sum of +powers of 2. +If the result loops with period = 1 with sign +, + you are on a sign-magnitude machine. +If the result loops with period = 1 at -1, + you are on a twos-complement machine. +If the result loops with period > 1, including the beginning, + you are on a ones-complement machine. +If the result loops with period > 1, not including the beginning, + your machine isn't binary -- the pattern should tell you + the base. +If you run out of memory, you are on a string or Bignum system. +If arithmetic overflow is a fatal error, some fascist pig with a + read-only mind is trying to enforce machine independence. + But the very ability to trap overflow is machine + dependent. +By this strategy, consider the universe, or, more precisely, +algebra: + let X = the sum of many powers of two = ...111111 + now add X to itself; X + X = ...111110 + thus, 2X = X - 1 so X = -1 + therefore algebra is run on a machine (the universe) + which is twos-complement. + +ITEM (Liknaitzky): +To subtract the right half of an accumulator from the left +(as in restarting an AOBJN counter): IMUL A,[377777,,1] + +ITEM (Mitchell): +To make an AOBJN pointer when the origin is fixed +and the length is a variable in A: + HRLOI A,-1(A) + EQVI A,ORIGIN + +ITEM (Freiberg): +If instead, A is a pointer to the last word + HRLOI A,-ORIGIN(A) + EQVI A,ORIGIN +Slightly faster: change the HRLOIs to MOVSIs and the +EQVI addresses to -ORIGIN-1. These two routines are +clearly adjustable for BLKOs and other fenceposts. + ITEM (Gosper, Salamin, Schroeppel): +A miniature (recursive) sine and cosine routine: +COS: FADR A,[1.57079632679] ;pi/2 +SIN: MOVM B,A ;argument in A + CAMG B,[.00017] ;<= (CUBEROOT 3) / 2^13 + POPJ P, ;sin X = X, within 27. bits + FDVRI A,(-3.0) + PUSHJ P,SIN ;sin -X/3 + FMPR B,B + FSC B,2 + FADRI B,(-3.0) + FMPRB A,B ;sin X = 4(sin -X/3)^3-3(sin -X/3) + POPJ P, ;sin in A, sin or ABS(sin) in B +;ABS(sin) in B occurs when angle is smaller than end test + +Changing both -3.0's to +3.0's gives sinh: +sinh X = 3 sinh X/3 + 4 (sinh X/3)^3. +Changing the first -3.0 to a +9.0, then inserting PUSHJ P,.+1 +after PUSHJ P,SIN gains about 20% in speed and uses +half the pushdown space (< 5 levels in the first 4 quadrants). +PUSHJ P,.+1 is a nice way to have something happen twice. +Other useful angle multiplying formulas are +tanh X = (2 tanh X/2)/(1 + (tanh X/2)^2) +tan X = (2 tan X/2)/(1 - (tan X/2)^2), if infinity is +handled correctly. For cos and cosh, one can use +cos X = 1 - 2 (sin X/2)^2, cosh X = 1 + 2 (sinh X/2)^2. + + X +In general, to compute functions like e , cos X, elliptic +functions, etc. by iterated application of double and triple +argument formulas, it is necessary to subtract out the constant +in the Taylor series and transform the range reduction formula +accordingly. Thus: + F(X) = cos(X)-1 F(2X) = 2F*(F+2) F(EPSILON) = -EPSILON^2/2 + X + G(X) = e -1 G(2X) = G*(G+2) G(EPSILON) = EPSILON + +This is to prevent the destruction of the information in the +range-reduced argument by the addition of a quantity near 1 upon +the success of the EPSILON test. The addition of such a quantity +in the actual recurrences is OK since the information is restored +by the multiply. In fact, a cheap and dirty test for F(EPSILON) +sufficiently small is to see if the addition step has no effect. +People lucky enough to have a square root instruction can get +natural log by iterating X_X/(SQRT(1+X) + 1) until 1+X = 1. +Then multiply by 2^(number of iterations). Here, a LSH or FSC would work. + +ITEM (Gosper, Schroeppel): +(Numbers herein are decimal.) +The correct epsilon test in such functions as the foregoing +SIN are generally the largest argument for which addition of +the second term has no effect on the first. In SIN, the +first term is x and the second is -x^3/6, so the answer is +roughly the x which makes the ratio of those terms 1/2^27; +so x = (SQRT 3) / 2^13. But this is not exact, since the precise +cutoff is where the neglected term is the power of 2 whose 1 bit +coincides with the first neglected (28th) bit of the fraction. +Thus, x^3/6 = 1/2^27 * 1/2^13, so x = (CUBEROOT 3) / 2^13. + ITEM (Gosper): +Here is a way to get log base 2. A and B are consecutive. +Call by PUSHJ P,LOG2 with a floating point argument in A. + +LOG2: LSHC A,-33 + MOVSI C,-201(A) + TLC C,211000 ;Speciner's bum + MOVEI A,200 ;exponent and sign sentinel +LOGL: LSH B,-9 +REPEAT 7, FMPR B,B ;moby flunderflo + LSH B,2 + LSHC A,7 + SOJG A,LOGL ;fails on 4th try + LSH A,-1 + FADR A,C + POPJ P, ;answer in A + +Basically, you just square seven times and use the low seven +bits of the exponent as the next seven bits of the log. + ITEM (Gosper): +To swap the contents of two locations in memory: + EXCH A,LOC1 + EXCH A,LOC2 + EXCH A,LOC1 +Note: LOC1 must not equal LOC2! If this can happen, +use MOVE-EXCH-MOVEM, clobbering A. + +ITEM (Gosper): +To swap two bits in an accumulator: + TRCE A,BITS + TRCE A,BITS + TRCE A,BITS +Note (Nelson): last TRCE never skips, and used to be +a TRC, but TRCE is less forgettable. Also, use TLCE +or TDCE if the bits are not in the right half. + +ITEM (Sussman): +To exchange two variables in LISP without using a third variable: + (SETQ X (PROG2 0 Y (SETQ Y X))) + +ITEM (Samson): +To take MAX in A of two byte pointers +(where A and B are consecutive accumulators): + + ROTC A,6 + CAMG A,B + EXCH A,B + ROTC A,-6 + +ITEM (Freiberg): +A byte pointer can be converted to a character address < 2^18. by +MULI A,<# bytes/word> followed by SUBI B,1-<# b/w>(A). +To get full word character address, use SUB into a magic table. + +ITEM (Gosper, Liknaitzky): +To rotate three consecutive accumulators N < 37. places: + + ROTC A,N + ROT B,-N + ROTC B,N + +Thus M AC's can be ROTC'ed in 2M-3 instructions. +(Stallman): For 73. > N > 35.: + + ROTC A,N-36. + EXCH A,C + ROT B,36.-N + ROTC A,N-72. + ITEM (Gosper, Freiberg): +;B gets 7 bit character in A with even parity + IMUL A,[2010040201] ;5 adjacent copies + AND A,[21042104377] ;every 4th bit of left 4 copies + right copy + IDIVI A,17_7 ;casting out 15.'s in hexadecimal shifted 7 + +;odd parity on 7 bits (Schroeppel) + IMUL A,[10040201] ;4 adjacent copies + IOR A,[7555555400] ;leaves every 3rd bit+offset+right copy + IDIVI A,9_7 ;powers of 2^3 are +-1 mod 9 +;changing 7555555400 to 27555555400 gives even parity + +;if A is a 9 bit quantity, B gets number of 1's (Schroeppel) + IMUL A,[1001001001] ;4 copies + AND A,[42104210421] ;every 4th bit + IDIVI A,17 ;casting out 15.'s in hexadecimal + +;if A is 6 bit quantity, B gets 6 bits reversed (Schroeppel) + IMUL A,[2020202] ;4 copies shifted + AND A,[104422010] ;where bits coincide with reverse repeated base 2^8 + IDIVI A,377 ;casting out 2^8-1's + +;reverse 7 bits (Schroeppel) + IMUL A,[10004002001] ;4 copies sep by 000's base 2 (may set arith. o'flow) + AND A,[210210210010] ;where bits coincide with reverse repeated base 2^8 + IDIVI A,377 ;casting out 377's + +;reverse 8 bits (Schroeppel) + MUL A,[100200401002] ;5 copies in A and B + AND B,[20420420020] ;where bits coincide with reverse repeated base 2^12 + ANDI A,41 ;" + DIVI A,1777 ;casting out 2^12-1's + ITEM (PDP-1 hackers): +foo, lat /DATAI switches + adm a /ADDB + and (707070 + adm b + iot 14 /output AC sign bit to a music flip-flop + jmp foo + +Makes startling chords, arpeggios, and slides, with just the +sign of the AC. This translates to the PDP-6 (roughly) as: + +FOO: DATAI 2 + ADDB 1,2 + AND 2,[707070707070] ;or 171717171717, 363636363636, 454545454545, ... + ADDB 2,3 + LDB 0,[360600,,2] + JRST FOO +Listen to the square waves from the low bits of 0. + +ITEM (in order of one-ups-manship: +Gosper, Mann, Lenard, [Root and Mann]): +To count the ones in a PDP-6/10 word: + + LDB B,[014300,,A] ;or MOVE B,A then LSH B,-1 + AND B,[333333,,333333] + SUB A,B + LSH B,-1 + AND B,[333333,,333333] + SUBB A,B ;each octal digit is replaced by number of 1's in it + LSH B,-3 + ADD A,B + AND A,[070707,,070707] + IDIVI A,77 ;casting out 63.'s + +These ten instructions, with constants extended, would work +on word lengths up to 62.; eleven suffice up to 254.. + ITEM (Jensen): +Useful strings of non-digits and zeros can arise when +carefully chosen negative numbers are fed to unsuspecting +decimal print routines. Different sets arise from +different methods of character-to-digit conversion. +Example (Gosper): + +DPT: IDIVI F,12 + HRLM G,(P) ;tuck remainder on pushdown list + SKIPE F + PUSHJ P,DPT + LDB G,[220600,,(P)] ;retrieve low 6 bits of remainder + TRCE G,"0 ;convert digit to character + SETOM CCT ;that was no digit! + +TYO: .IOT TYOCHN,G ;or DATAO or IDPB ... + AOS G,CCT + POPJ P, + +This is the standard recursive decimal print of the positive +number in F, but with a LDB instead of a HLRZ. +It falls into the typeout routine which returns +in G the number of characters since the last carriage return. +When called with a -36., DPT types carriage return, line feed, +and resets CCT, the character position counter. + +ITEM (Gosper): +Since integer division can never produce a larger quotient than +dividend, doubling the dividend and divisor beforehand will +distinguish division by zero from division by 1 or anything +else, in situations where division by zero does nothing. + +ITEM (Gosper): +The fundamental operation for building list structure, called +CONS, is defined to: find a free cell in memory, store the +argument in it, remove it from the set of free cells, return +a pointer to it, and call the garbage collector when the set +is empty. This can be done in two instructions: + +CONS: EXCH A,[EXCH A,[...[PUSHJ P,GC]...]] + EXCH A,CONS + +Of course, the address-linked chain of EXCH's indicated by +the nested brackets is concocted by the garbage collector. +This method has the additional advantage of not constraining +an accumulator for the free storage pointer. + +UNCONS: HRLI A,(EXCH A,) + EXCH A,CONS + EXCH A,@CONS + +Returns cell addressed by A to free storage list; +returns former cell contents in A. + ITEM (Gosper): +The incantation to fix a floating number is usually + MULI A,400 ;exponent to A, fraction to A+1 + TSC A,A ;1's complement magnitude of excess 200 exponent + ASH A+1,-200-27.-8(A) ;answer in A+1 +If number is known positive, you can omit the TSC. +On the PDP-10 + UFA A,[+or-233000,,] ;not in PDP-6 repertoire + TLC A+1,233000 ;if those bits really bother you + +When you know the sign of A, and ABS(A) < 2^26, you can + FAD A,[+or-233400,,] ;or FADR for rounded fix! + TLC A,233400 ;if those bits are relevant +where the sign of the constant must match A's. +This works on both machines and doesn't involve A+1. +On the 10, FADRI saves a cycle and a constant, and rounds. + +ITEM (Gosper, Nelson): +21963283741.=243507216435 is a fixed point of the float +function on the PDP-6/10, i.e., it is the only positive +number whose floating point representation equals its fixed. + +ITEM (Gosper): +To get the next higher number (in A) with the same number +of 1 bits: (A, B, C, D do not have to be consecutive) + MOVE B,A + MOVN C,B + AND C,B + ADD A,C + MOVE D,A + XOR D,B + LSH D,-2 + IDIVM D,C + IOR A,C + ********************************************* +PROGRAMMING ALGORITHMS, HEURISTICS +********************************************* + +ITEM (Gosper): +The "banana phenomenon" was encountered when processing a +character string by taking the last 3 letters typed out, +searching for a random occurrence of that sequence in the text, +taking the letter following that occurrence, typing it out, and +iterating. This ensures that every 4-letter string output occurs +in the original. The program typed BANANANANANANANA.... +We note an ambiguity in the phrase, "the Nth occurrence of." +In one sense, there are five 00's in 0000000000; in another, +there are nine. The editing program TECO finds five. Thus it +finds only the first ANA in BANANA, and is thus obligated to type +N next. By Murphy's Law, there is but one NAN, thus forcing A, +and thus a loop. An option to find overlapped instances would be +useful, although it would require backing up N-1 characters +before seeking the next N character string. + +ITEM (Gosper): DRAWING CURVES INCREMENTALLY + + Certain plotters and displays are constrained to +approximate curves by a sequence of king-moves between points +on a lattice. + + Many curves and contours are definable by F(X,Y) = 0 with +F changing sign on opposite sides of the curve. The following +algorithm will draw most such curves more accurately than +polygonal approximations and more easily than techniques which +search for a "next" X and Y just one move away. + + We observe that a good choice of lattice points is just +those for which F, when evaluated on one of them, has opposite +sign and smaller magnitude than on one or more of its four +immediate neighbors.* This tends to choose the nearer endpoint +of each graph paper line segment which the curve crosses, if near +the curve F is monotone with distance from the curve. + + First, divide the curve into arcs within which the +curve's tangent lies within one 45 degree semiquadrant. We can +show that for reasonable F, only two different increments (say +north and northwest) are needed to visit the desired points. + + Thus, we will be changing one coordinate (incrementing Y) +every step, and we have only to check whether changing the other +(decrementing X) will reduce the magnitude of F. (If F increases +with Y, F(X,Y+1) > -F(X-1,Y+1) means decrement X.) F can often +be manipulated so that the inequality simplifies and so that F is +easily computed incrementally from X and Y. + As an example, the following computes the first +semiquadrant of the circle + + F = X^2+Y^2-R^2 = 0. + +C0: F_0, Y_0, X_R + +C1: F_F+2Y+1, Y_Y+1 + +C2: if* F >= X, F_F-2X+1, X_X-1 + +C3: if Y < X-1, go to C1 + +C4: (Link to next arc) if Y = X-1, Y_Y+1, X_X-1 + + This can be bummed by maintaining Z = 2Y+1 instead of Y. +Symmetry may be used to compute all eight semiquadrants at once, +or the loop may be closed at C2 and C3 with two PUSHJ's to +provide the palindrome of decisions for the first quadrant. +There is an expression for the number of steps per quadrant, +but it has a three-way conditional dependent upon the midpoint +geometry. Knowing this value, however, we can replace C3 and C4 +with a simple loop count and an odd-even test for C4. + + The loop must be top-tested (C3 before C1) if the +"circle" R = 1, with four diagonal segments, is possible. + + All this suggests that displays might be designed with an +increment mode which accepts bit strings along with declarations +of the form: "0 means north, 1 means northwest". 1100 (or 0011) +will not occur with a curve of limited curvature; thus, it could +be used as an escape code, but this would be an annoying +restriction. + +*In case of a tie, i.e., F has equal magnitudes with opposite +signs on adjacent points, do not choose both points but rather +have some arbitrary yet consistent preference for, say, the outer +one. The problem can't arise for C2 in the example because the +inequality F >= X is really F > -(F-2X+1) or F > X-.5. + ITEM (Schroeppel, Salamin): +Suppose Y satisfies a differential equation of the form + P(X)Y(Nth derivative) + ..... + Q(X) = R(X) +where P, ..... Q, and R are polynomials in X +(for example, Bessel's equation, X^2Y''+XY'+(X^2-N^2)Y = 0) +and A is an algebraic number. Then Y(A) can be evaluated to +N places in time proportional to N(ln N)^3. + X +Further, e and ln X or any elementary function can be evaluated +to N places in N(ln N)^2 for X a real number. If F(X) can be +evaluated in such time, so can the inverse of F(X) (by Newton's +method), and the first derivative of F(X). Also, ZETA(3) and GAMMA +can be done in N(ln N)^3. + +ITEM (Gosper): +A program which searches a character string for a given +substring can always be written by iterating the sequence +fetch-compare-transfer (ILDB-CAIE-JRST on the PDP6/10) once +for each character in the sought string. The destinations +of the transfers (address fields of the JRST's) must, however, +be computed as functions of the sought string. +Let + 0 1 2 3 4 + S A S S Y + 0 1 0 2 2 +stand for the program + +T0: ILDB C,A ;C gets next char from pointer in A +T1: CAIE C,"S ;skip if it's an S + JRST T0 ;loop back on failure + ILDB C,A ;next +T2: CAIE C,"A ;skip if A + JRST T1 ;could be an S + ILDB C,A +T3: CAIE C,"S + JRST T0 ;S, A, non S, so start over + ILDB C,A ;next +T4: CAIE C,"S + JRST T2 ;could be SAS.ASSY + ILDB C,A + CAIE C,"Y + JRST T2 ;could be SASS.ASSY +;found SASSY + +In other words, a number > 0 in the top row is a location +in the program where the corresponding letter of the +middle row is compared with a character of the input string. +If it differs, the number in the bottom row indicates the +location where comparison is to resume. If it matches, +the next character of the middle row is compared with the +next character of the input string. + Let J be a number in the top row and K be the number +below J, so that TK is the address field of the Jth JRST. +For each J = 1, 2, ... we compute K(J) as follows: +K(1) = 0. Let P be a counter, initially 0. +For each succeeding J, increment P. If the Pth letter = the Jth, +K(J) = K(P). Otherwise, K(J) = P, and P is reset to 0. (P(J) +is the largest number such that the first P characters match +the last P characters in the first J characters of the sought +string.) + +J= 0 1 0 1 2 3 4 5 + M I S S I S S I P P I I S S I S S I P P I +K(J)= 0 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 5 1 0 + + 0 1 2 3 0 1 2 3 + C O C A C O L A S A S S A F R A S + 0 1 0 2 0 1 3 1 0 1 0 2 1 3 1 1 0 + +To generalize this method to search for N strings at once, we +produce a program of ILDB-CAIE-JRST's for each of the sought +strings, omitting the initial ILDB from all but the first. We +must compute the destination of the Jth JRST in the Ith program, +TKM(I,J), which is the location of the Kth compare in the Mth +program. + +It might be reasonable to compile such an instruction sequence +whenever a search is initiated, since alternative schemes usually +require saving or backing up the character pointer. + +ITEM (Gosper): +A problem which may arise in machine processing of visual +information is the identification of corners on a noisy boundary +of a polygon. Assume you have a broken line. If it is a closed +loop, find the vertex furthest from the centroid (or any place). +Open the loop by making this place both endpoints and calling it +a corner. We define the corner of a broken line segment to be +the point the sum of whose distances from the endpoints is +maximal. This will divide the segment in two, allowing us to +proceed recursively, until our corner isn't much cornerier than +the others along the line. + +The perpendicular distance which the vector C lies +from the line connecting vectors A and B is just + (C - A) CROSS (B - A) + ----------------------- , + 2 ABS(A - B) +but maximizing this can lose on very pointy V's. +The distance sum hack can lose on very squashed Z's. + ********************************************* +HARDWARE +********************************************* + +ITEM (Gosper): + A bug you might try to avoid when designing floating +point hardware, relating to excess-200, 1's complement exponent, +2's complement fraction convention: + +1) An advantage is that negation and numerical comparison can +be accomplished with the same instructions for both fixed and +floating point numbers. + +2) A disadvantage is that the termination of the +normalization process is ambiguous. Normally, when +the sign bit unequals the highest bit of fraction, +the number is normalized. A special case arises with + n -n +negated powers of two. (That is, -(2 ), not (2) .) +Then the fraction is 400,,0 and the sign is - also. + + This means it is necessary to check whether shifting left +one more bit will bring in a one: + if it brings in a zero, you will over-normalize + if it brings in a one, you should do it +If you should but don't, rounding will un-normalize, and when you +then re-normalize, the normalizing amount will be doubled, so you +will be off by 2 smidgens (that is, the next to low order bit). +Note that rounding can over-normalize as well as un-normalize, +so you can't just stop normalization after rounding. + + You might check this in your PDP-6/10. For example, +combine 201400,,0+EPS with minus 200777,,777777+2EPS. For +0 =< EPS =< 7777, the correct FMP result is minus 200777,,777776, +and the correct FMPR result is minus 200777,,777777. + + Over-normalized negative powers of 2 work in compares and +most floating arithmetic. They lose with MOVN and as dividends. +Unnormalized floating operands win completely on the PDP-10, +except as divisors and dividends, the latter suffering truncation +error. + ITEM (Roe): VOLTAGE REGULATORS +Fairchild is now supplying positive voltage +regulators costing about 2 dollars in lots +of 1 (for example, the uA7805 for +5 volts). + +ITEM (Roe): CURRENT MIRRORS +The CA3083 (and CA3084) transistor arrays can be used to make +neat current mirrors. (A current mirror supplies a current on +one wire equal to that drawn from a second wire.) + +ITEM (Roe): ONE-SHOT +A dual MOS D-type flip-flop (such as the CD4013AE) +can be used to make a one-shot as follows: + +ITEM (Roe): OSCILLATORS +Everyone has their own favorite oscillator circuits; +here are some we like. +I crystal, overtone, transistor +II crystal, fundamental, transistor (drives at least 1 TTL load) +III crystal, fundamental, CMOS, low frequency (drives 1 TTL load; +at 5.4 volts and no load, draws 330 microamperes; with a 165 KHz, +32 pf crystal, varies about 10 Hz per volt of Vcc) +IV crystal, fundamental, IC (a favorite of Nelson's, but be +careful and lucky or it may oscillate at a frequency determined +by the crystal holder capacitance and not by the crystal; note +similarity to non-crystal oscillator V) +V not crystal controlled; for comparison with IV +VI The following blocking oscillator is quite uncritical of +component values, with the exception that the turns ratio be +such that -Vb (see graph) not exceed BVebo (about 5 volts for +silicon transistors). + +ITEM (Roe): FM RADIO LINK +In work on education at our lab, we built a motorized "turtle" +controlled by computer commands in the child-oriented language +"Logo." The following is a transmitter designed as a radio link +between the computer and turtle. Input (modulation) is either +0 or +12 volts; output is about 88MHz. Use a commercial FM tuner +as receiver. Note: this transmitter is ILLEGAL no matter what; +part 15 low power rule only allows if duty is less than about +1 second per 15 minutes. Don't worry about it unless you +interfere with broadcast stations. + ITEM (Roe): PHONE LINE XMTR, RCVR +When the chess program written at our lab is playing in a chess +tournament, a human attendant at the tournament moves the pieces, +punches the clock, and communicates with the program via a +portable terminal coupled to a telephone line. It is desirable +that the program know when its chess clock is running, even +though the attendant may not notice immediately that the opponent +has made his move and punched the clock. Therefore we built a +clock holder with a microswitch to sense the clock state. +The following is a 10 mw transmitter whose input is the +microswitch and whose output goes onto the phone line. +It switches between two frequencies, about 320 and 470 Hz. +Also shown is the receiver. Input should be at least 100 mv rms +(threshold is 20 mv and overload is above 68 volts) with peak to +peak signal to noise ratio greater than 4:1. As we all know, +connections to phone lines are illegal unless made through a data +coupler supplied by TPC (The Phone Company). + +ITEM (Roe): DC MOTOR VELOCITY SERVO +One version of the "turtle" mentioned above (see RADIO LINK) uses +a DC motor to drive each of its two powered wheels. Since its +path is to be as straight as possible, a triangular pulse is +generated (to represent one "step" of the motor) and the motor's +velocity servoed to this analog command. An additional digital +command enables foreward or reverse motion. Diagram I shows a +simplified velocity servoing circuit. It has the disadvantage +that only half the maximum voltage available (-V to +V) can be +applied across the motor at any one time. Diagram II shows the +actual circuit used in the turtle. + +ITEM (Roe): OPTICAL COUPLER +When two circuits are at potentials differing by a few hundred +volts but wish to communicate with each other, one solution is +to use an optical coupler. These employ a light-emitting device +placed close to a light-sensitive device. Diodes make very +fast-responding sensors, but the signal from a light-sensitive +transistor is much stronger. Shown is a compromise, using a +transistor as a diode, with associated cleverness to get the +delay (from input to output) down from 10 microseconds to 1. + +ITEM (Roe): PHOTOCATHODE CURRENT OSCILLATOR +In our fourth computer-interfaced image sensing device, TVD +(really a vidissector, not a TV), the photocathode sits at +several thousand volts negative. Nevertheless, one wishes to +sense the current it draws, since overcurrent should shut down +the photocathode voltage to avoid damage to the photocathode. +The following circuit draws no more than 400 microamperes at +10 volts (at 20 KHz out; about 200 microamperes at 10 KHz) and +couples the current information out as the frequency sent to T2, +whose coils are wound on opposite sides of a ceramic ferrite. + ITEM (Roe): DEFLECTION AMPLIFIER +TVD, mentioned above, uses a very carefully designed printed +circuit amplifier to supply current to its magnetic deflection +coils. Except for the notes with the diagram, we submit it +without further explanation or cautions. + +Notes: +1 Except where noted, resistors 10%, 1/4 watt. +2 Capacitances in microfarads/volts; electrolytics aluminum. +3 Diodes 1N4727, 1N4154,1N4009 etc.; stored charge no more than + 80 picocoulombs at 1 milliampere foreward current. +4 1D103 = GE thermistor mounted at center of main heat sink. +5 220J = Analog Devices chopper amplifier. +6 * = temperature protection circuit (overtemperature cutout). +7 Q2, Q3, Q4, Q5, Q6, Q12, Q13, Q14, Q15, Q16 mounted + on one 1 Centigrade degree per watt heat sink (e.g. + Wakefield 621K 1/2 inch in front of Rotron Muffin fan). + Case temperature about 70 degrees C max. + Ground heat sink and insulate transistors. +8 All transistors Motorola. +9 All zeners 1 watt. +10 VE48X = Varo; could be two 2 A 50 PIV fast recovery. +11 Output capacitance about 800 pf; damping R about 150 ohms + for critical damping. +12 Slews from + (or -) 2 A to - (or +) 2 A in 4 microseconds; + dE/dt at hot side of deflection coil is about a billion v/sec. +13 Layout is critical, as with most fast high-gain circuits. + A By-passing and lead inductance: Short wide strips (or, + better, a ground plane) should be used for ground bus, + and ceramic capacitors with leads as short as practicable + used for bypassing. Best bypass capacitor is + Allen-Bradley CL series. + B Ground loops: reference ground (triangles) and power ground + must be interconnected only at the cold side of the sense + resistor; take care to avoid stray current through the cold + side of the signal input. + C In general, the device should be constructed like + a 144 MHz transmitter to avoid its becomming one. +14 The 100 pf stabilizing capacitor may want to be higher to +decrease hunting and ringing, which could improve settling time +more than the reduced gain-bandwidth would increase it. + +Q1, Q12, Q13 MPS-U01 +Q11, Q2, Q3 MPS-U51 +Q4, Q5, Q6 2N5194 +Q14, Q15, Q16 2N5191 +Q7 MPS-U02 +Q17 MPS-U52 +Q8, Q19 2N3906 +Q9, Q18 2N3904 +  \ No newline at end of file diff --git a/doc/mb/hakmem.17 b/doc/mb/hakmem.17 new file mode 100644 index 00000000..6f06eb7d --- /dev/null +++ b/doc/mb/hakmem.17 @@ -0,0 +1,1029 @@ +********************************************* +GEOMETRY, ALGEBRA, CALCULUS +********************************************* + +ITEM (SCHROEPPEL): +FUNCTIONS OF THE FORM AY'' + BY' + CY = D +WHERE A, B, C, AND D ARE FUNCTIONS OF X +CAN BE CONVERTED TO THE FORM W' = W^2 - E +WHERE E = B^2/4A^2 - C/A + (AB' - BA')/2A^2. +IN ADDITION, A PARTICULAR SOLUTION OF THE NEW FORM +IS ENOUGH BECAUSE SOLUTIONS DIFFER BY A SOLVABLE AMOUNT. +FOR EXAMPLE, BESSEL'S FUNCTION BECOMES + W' = W^2 + 1 - (N^2 - 1/4)/X^2 +SIMILARLY, Y''' + AY'' + BY' + CY = D +CAN BE REDUCED TO + Y'' = Y^2 + A AND + Y'' + 3YY' + Y^3 + AY + B = 0 + +ITEM (SCHROEPPEL): +GAMMA(1/3) AND GAMMA(2/3) ARE INTEREXPRESSIBLE. +GAMMA(1/4) AND GAMMA(3/4) ARE INTEREXPRESSIBLE. +THUS THESE TWO PAIRS ARE OF DIMENSIONALITY ONE. +GAMMA(1/10) AND GAMMA(2/10) ARE SUFFICIENT TO EXPRESS GAMMA(N/10) FOR ALL N. +GAMMA(1/12) AND GAMMA(2/12) ARE SUFFICIENT TO EXPRESS GAMMA(N/12) FOR ALL N. +GAMMA(1/3) AND GAMMA(1/4) ARE SUFFICIENT TO EXPRESS GAMMA(N/12) FOR ALL N. +THUS THE THREE CASES ABOVE ARE OF DIMENSIONALITY TWO. +PROBLEM: FIND SOME ORDER TO THIS DIMENSIONALITY BUSINESS. + +ITEM (JAN KOK): +PROBLEM: GIVEN A REGULAR N-GON WITH ALL DIAGONALS DRAWN, +HOW MANY REGIONS ARE THERE? IN PARTICULAR, HOW MANY TRIPLE (OR N-TUPLE) +CONCURRENCES OF DIAGONALS ARE THERE? + +ITEM (SCHROEPPEL): +REGARDING CONVERGENCE OF NEWTON'S METHOD FOR QUADRATIC EQUATIONS: +DRAW THE PERPENDICULAR BISECTOR OF THE LINE CONNECTING THE TWO ROOTS. +POINTS ON EITHER SIDE CONVERGE TO THE CLOSEST ROOT. ON THE LINE: +1 THEY DO NOT CONVERGE +2 THERE IS A DENSE SET OF POINTS WHICH INVOLVE DIVISION BY ZERO +3 THERE IS A DENSE SET OF POINTS WHICH LOOP, BUT ROUNDOFF ERROR + PROPAGATES SO ALL LOOPS ARE UNSTABLE +4 BEING ON THE LINE IS ALSO UNSTABLE (UNLESS, OF COURSE, THE ROOTS ARE + IMAGINARY AND YOU ARE ON THE REAL AXIS) + +ITEM (SCHROEPPEL): +THE MOST PROBABLE SUIT DISTRIBUTION IN BRIDGE HANDS IS 4-4-3-2, AS COMPARED +TO 4-3-3-3, WHICH IS THE MOST EVENLY DISTRIBUTED. THIS IS BECAUSE THE +WORLD TENDS TO HAVE UNEQUAL NUMBERS, WHICH IS SORT OF A THERMODYNAMIC EFFECT +SAYING THINGS WILL NOT BE IN THE STATE OF LOWEST ENERGY, BUT IN THE STATE +OF LOWEST DISORDERED ENERGY. + +ITEM (SCHROEPPEL): +BY MATHLAB, THE DISCRIMINANT OF X^4 + FX^3 + GX^2 + HX + I IS +(AS THE DISCRIMINANT OF AX^2 + BX + C IS B^2 - 4AC): + - 27 H^4 + 18 FGH^3 - 4 F^3H^3 - 4 G^3H^2 + F^2G^2H^2 + + I * [144 GH^2 - 6 F^2H^2 - 80 FG^2H + 18 F^3GH + 16 G^4 - 4 F^2G^3] + + I^2 * [- 192 FH - 128 G^2 + 144 F^2G - 27 F^4] + - 256 I^3 + +ITEM (SCHROEPPEL): +IF A IS THE FIRST SYMMETRIC FUNCTION OF N VARIABLES = X + Y + Z + ... +AND B IS THE SECOND SYMMETRIC FUNCTION OF N VARIABLES = XY + XZ + ... + YZ + ... +(B = SUM OF PAIRS), THEN X^2 + Y^2 + Z^2 + ... = A^2 - 2B. +FURTHERMORE, SIMILAR RULES EXIST FOR SUMS OF OTHER POWERS, SUCH AS +X^3 + Y^3 + Z^3 + ... AND X^4 + Y^4 + Z^4 + ..., ETC. +FOR EXAMPLE, X^4 + Y^4 + Z^4 + ... = A^4 - 4A^2B + 2B^2 + 4AC - 4D. + +ITEM (SCHROEPPEL): +IF X^4 + BX^2 + CX + D = 0, THEN +2X = +OR- (SQRT Z1) +OR- (SQRT Z2) +OR- (SQRT Z3) WHERE AN ODD NUMBER OF SIGNS ARE PLUS. +Z1, Z2, Z3 ARE ROOTS OF Z^3 - 2BZ^2 + (B^2 - 4D)Z - C^2 = 0. +PROBLEM: REPLACING C BY -C SHOULD TAKE X INTO -X, BUT THE CUBIC WILL HAVE THE SAME ROOTS. +MAYBE THERE IS AN ERROR, OR MAYBE YOU TAKE AN EVEN NUMBER OF SIGNS +TO BE PLUS? + +ITEM (SCHROEPPEL): +SOLUTIONS TO F(X) = X^3 - 3BX^2 + CX + D = 0 ARE +B - K * CUBE ROOT [F(B)/2 + SQRT [(F(B)/2)^2 + (F'(B)/3)^3]] + - L * CUBE ROOT [F(B)/2 - SQRT [(F(B)/2)^2 + (F'(B)/3)^3]] +WHERE OMEGA = CUBE ROOT [1], +K = L = 1 GIVES THE FIRST ROOT +K = OMEGA, L = OMEGA^2 GIVES THE SECOND ROOT +K = OMEGA^2, L = OMEGA GIVES THE THIRD ROOT + +ITEM (GOSPER): +THE MYTH THAT ANY GIVEN PROGRAMMING LANGUAGE IS MACHINE INDEPENDENT +IS EASILY EXPLODED BY COMPUTING THE SUM OF POWERS OF TWO. +IF THE RESULT LOOPS WITH PERIOD = 1 WITH SIGN +, YOU ARE ON A SIGN-MAGNITUDE MACHINE. +IF THE RESULT LOOPS WITH PERIOD = 1 WITH SIGN -, YOU ARE ON A TWOS-COMPLEMENT MACHINE. +IF THE RESULT LOOPS WITH PERIOD > 1, INCLUDING THE BEGINNING, YOU ARE ON A ONES-COMPLEMENT MACHINE. +IF THE RESULT LOOPS WITH PERIOD > 1, NOT INCLUDING THE BEGINNING, YOU ARE ON A DECIMAL MACHINE. +IF YOU RUN OUT OF MEMORY, YOU ARE ON A STRING SYSTEM/MACHINE. +BY THIS STRATEGY, CONSIDER THE UNIVERSE, OR, MORE PRECISELY, ALGEBRA: + LET X = THE SUM OF MANY POWERS OF TWO = ...111111 + NOW ADD X TO ITSELF; X + X = ...111110 + THUS, 2X = X - 1 SO X = -1 + THEREFORE ALGEBRA IS RUN ON A MACHINE (THE UNIVERSE) WHICH IS TWOS-COMPLEMENT. + +ITEM (SALAMIN): +CONSTRUCT CONFORMAL MAPPINGS OF EUCLIDEAN N-SPACE INTO ITSELF BY +STEREOGRAPHIC PROJECTION ONTO A N-SPHERE, ROTATING THE SPHERE, AND +PROJECTING BACK. +PROBLEM: GENERALIZE HOMOGRAPHIC TRANSFORMATIONS AND ANALYTIC FUNCTIONS. + +ITEM (GOSPER): +PROBLEM: PICK A RANDOM ANGLE. TURN THAT ANGLE AND GO A UNIT STEP. +TURN TWICE THE ANGLE AND GO A UNIT STEP. TURN FOUR TIMES ... +FOR WHAT INITIAL ANGLES IS YOUR LOCUS BOUNDED? + ********************************************* +BOOLEAN ALGEBRA +********************************************* + +ITEM (SCHROEPPEL): +IN AN ATTEMPT TO GET A HOLD ON THE PROBLEM OF MINIMAL LOGIC GATE SYNTHESIS, +WHICH MAY (HOPEFULLY) LEAD TO USEFUL ALGORITHMS, SOME PARTIAL RESULTS ARE: +DESCRIPTION OF RESTRICTIONS: + 2 BINARY INPUTS + OUTPUTS COMPOSED OF 0, 1, OR DON'T CARE +FOR FUNCTIONS OF 2 INPUT BITS, A MINIMAL SYNTHESIS EXISTS. +RULES TO EVALUATE "MINIMAL": +1 -- NOT FUNCTIONS ARE FOR FREE +2 -- CHARGE 1 FOR EACH AND GATE OF 2 INPUTS +3 -- THERE IS NO FEEDBACK +4 -- AN XOR GATE COSTS (I.E., CAN BE MADE OF NO FEWER THAN) 3 AND GATES +5 -- A MAJORITY OF 3 FUNCTION COSTS 4 AND GATES +EXAMPLE: PRIME PREDICATE ON 4 BITS COSTS AT MOST 7 + +ITEM (SPECINER): +N NUMBER OF BOOLEAN MONOTONIC FUNCTIONS OF N VARIABLES +0 2 (T, F) +1 3 (T, F, P) +2 6 (T, F, P, Q, P AND Q, P OR Q) +3 20 +4 168 +5 7581 +6 7,828,354 + +ITEM (SCHROEPPEL): +APPLICABLE TO 2-NOTS PROBLEM (SYNTHESIZE A BLACK BOX WHICH COMPUTES +NOT-A, NOT-B, AND NOT-C FROM A, B, AND C, USING ONLY 2 NOTS): +FUNCTIONS SYNTHESIZABLE WITH ONE NOT ARE THOSE WHERE ANY UPWARD PATH +HAS AT MOST ONE DECREASE (THAT IS, FROM T TO F). + +ITEM (ROGER BANKS): +A VENN DIAGRAM FOR N VARIABLES WHERE THE SHAPE REPRESENTING EACH VARIABLE IS CONVEX +CAN BE MADE BY SUPERIMPOSING SUCCESSIVE M-GONS (M = 2, 4, 8, ...), +EVERY OTHER SIDE OF WHICH HAS BEEN PUSHED OUT TO THE CIRCUMSCRIBING CIRCLE. +THE NESTED M-GONS MUST SHRINK SLIGHTLY; PROBABLY RADIUS = 2, 3/2, 5/4, ETC. +IS SUFFICIENT. + +ITEM (SCHROEPPEL & WALTZ): +PROBLEM: COVER THE EXECUPORT CHARACTER RASTER COMPLETELY WITH THE MINIMUM NUMBER +OF CHARACTERS. THE THREE CHARACTERS I, H AND # WORKS. USING CAPITAL LETTERS ONLY, +THE FIVE CHARACTERS B, I, M, V AND X IS A MINIMAL SOLUTION. +FIND A GENERAL METHOD OF SOLVING SUCH PROBLEMS. + +ITEM (GOSPER): +PROBLEM: GIVEN SEVERAL BINARY NUMBERS, HOW CAN ONE FIND A MASK +WITH A MINIMAL NUMBER OF BITS ON WHICH, AND-ED WITH +EACH OF THE ORIGINAL NUMBERS, WILL PRESERVE THEIR DISTINCTNESS FROM EACH OTHER? + ********************************************* +RANDOM NUMBERS +********************************************* + +ITEM (SCHROEPPEL): +RANDOM NUMBER GENERATORS, SUCH AS ROLLO SILVER'S, WHICH USE SHIFTS AND XORS, +AND GIVE AS VALUES ONLY SOME PART OF THEIR INTERNAL STATE, CAN BE DESCRIBED +AND THEIR OPERATIONS INVERTED AND NUMBERS FROM THEM USED TO OBTAIN THEIR +TOTAL INTERNAL STATE. FOR EXAMPLE, 2 CONSECUTIVE VALUES FROM ROLLO'S +SUFFICE TO ALLOW PREDICTION OF ITS ENTIRE FUTURE. ROLLO'S IS: + REGISTER A GETS LOADED WITH "HIGH" WORD + REGISTER B GETS LOADED WITH "LOW" WORD + REGISTER A GETS STORED IN "LOW" WORD + PDP-6/10 INSTRUCTION LSHC A,35. IS EXECUTED + REGISTER A XORED WITH "HIGH" WORD GETS STORED IN "HIGH" AND IS VALUE +THIS SUGGESTS A SUSCEPTIBILITY TO ANALYSIS OF MECHANICAL CODE MACHINES. + +ITEM (SCHROEPPEL): +PROBLEM: WRITE A NUMBER WHICH HAS THE SAME NUMBER OF LETTERS AS THE NUMBER. +FOUR IS THE ONLY SOLUTION. + ********************************************* +NUMBER THEORY, PRIMES, PROBABILITY +********************************************* + +ITEM (SCHROEPPEL): +AFTER ABOUT 40 MINUTES OF RUN TIME TO VERIFY THE ABSENCE OF ANY +FURTHER FACTORS LESS THAN A PDP-10 WORD SIZE, +THE 125TH MERSENNE NUMBER, 2^125 - 1, WAS FACTORED ON TUESDAY, +JANUARY 5, 1971, IN 371 SECONDS RUN TIME AS FOLLOWS: +2^125 - 1 = 31 * 601 * 1,801 * 269,089,806,001 * 4,710,883,168,879,506,001 + +ITEM (SCHROEPPEL): +FOR A RANDOM NUMBER X, THE PROBABILITY OF ITS LARGEST PRIME FACTOR BEING +(1) GREATER THAN THE SQUARE ROOT OF X IS LN 2. +(2) LESS THAN THE CUBE ROOT OF X IS ABOUT 4.86%. +THIS SUGGESTS THAT SIMILAR PROBABILITIES ARE INDEPENDENT OF X; +FOR INSTANCE, THE PROBABILITY THAT THE LARGEST PRIME FACTOR OF X +IS LESS THAN THE 20TH ROOT OF X MAY BE A FRACTION INDEPENDENT OF X. +RELEVANT DATA: +RANGE COUNT CUM. SUM OF COUNT +10^12 TO 10^6 7198 [6944] 10018 +10^6 TO 10^4 2466 2820 +10^4 TO 10^3 354 402 [487] +10^3 TO 252 40 48 (252 = 10^(12/5)) +252 TO 100 7 8 +100 TO 52 1 1 +51 TO 1 0 0 +WHERE: +"COUNT" IS THE NUMBER OF NUMBERS WHOSE LARGEST PRIME FACTOR IS IN "RANGE" + FOR THE NUMBERS 10^12 + 1 TO 10^12 + 10018. +[ ] DENOTE THE EXPECTED VALUE OF ADJACENT ENTRIES. +THE NUMBER OF PRIMES IN 10^12 + 1 TO 10^12 + 10018 IS 335; + THE PRIME NUMBER THEOREM PREDICTS 363 IN THIS RANGE. +THIS IS RELEVANT TO KNUTH'S DISCUSSION OF FACTORING, VOL. 2, P. 351-354. + +ITEM (SCHROEPPEL): +TWIN PRIMES: + 166,666,666,667 = (10^12 + 2)/6 + 166,666,666,669 +THE PRIMES WHICH BRACKET 10^12 ARE 10^12 + 39 AND 10^12 - 11 +THE PRIMES WHICH BRACKET 10^15 ARE 10^15 + 37 AND 10^15 - 11 +THE NUMBER 23,333,333,333 IS PRIME. +THE NUMBER 166,666,666,666,667 IS PRIME, BUT + 166,666,666,666,669 IS NOT. +VARIOUS PRIMES, USING T = 10^12, ARE: + 40T + 1 + 62.5T + 1 + 500T - 1 + 500T - 7 + 200T - 3 + +ITEM (SCHROEPPEL): +RAMANUJAN'S PROBLEM OF SOLUTIONS TO 2^N - 7 = X^2 WAS SEARCHED TO +ABOUT N = 10^40; ONLY HIS SOLUTIONS (N = 3, 4, 5, 7, 15) WERE FOUND. + +ITEM (SCHROEPPEL): +TAKE A RANDOM REAL NUMBER AND RAISE IT TO LARGE POWERS; WE EXPECT THE +FRACTION PART TO BE UNIFORMLY DISTRIBUTED. +SOME EXCEPTIONS: +1 -- PHI = (1 + SQRT 5)/2 +2 -- ALL -1 < X < 1 +3 -- SQRT 2 (HALF ARE INTEGERS, OTHER HALF ARE UNCLEAR) +4 -- 1 + SQRT 2 -- PROOF: + (1 + SQRT 2)^N + (1 - SQRT 2)^N = INTEGER (BY INDUCTION); + THE (1 - SQRT 2)^N GOES TO ZERO. +5 -- 2 + SQRT 2 -- SIMILAR TO 1 + SQRT 2 +NOW, 3 + SQRT 2 IS SUSPICIOUS; IT LOOKS NON-UNIFORM, AND SEEMS TO +HAVE A CLUSTER POINT AT ZERO. PROBLEM: IS IT NON-UNIFORM? + +ITEM (SCHROEPPEL): +NUMBERS WHOSE RIGHT DIGIT CAN BE REPEATEDLY REMOVED AND THEY ARE STILL PRIME: +CONJECTURE: THERE ARE A FINITE NUMBER OF THEM IN ANY RADIX. +IN DECIMAL THERE ARE 51, THE LONGEST BEING 1,979,339,333 AND 1,979,339,339. + +ITEM (SCHROEPPEL): +PROBLEM: CAN EVERY POSITIVE INTEGER BE EXPRESSED IN TERMS OF 3 AND THE +OPERATIONS FACTORIAL AND INTEGER SQUARE ROOT? + +ITEM (SCHUTZENBERGER): +PROBLEM: USING N DIGITS, CONSTRUCT A STRING OF DIGITS WHICH AT NO TIME +HAS ANY SEGMENT APPEARING CONSECUTIVELY TWICE. +N = 2 => FINITE MAXIMUM STRING +N = 10 => KNOWN INFINITE +DETERMINE MAXIMUM STRING LENGTH FOR N = 3. +SUB-PROBLEM: HOW MANY SEQUENCES EXIST OF ANY PARTICULAR LENGTH? + +ITEM (BEELER): +THERE ARE NO ZEROS IN THE DECIMAL EXPRESSION OF POWERS OF TWO GREATER THAN 2^86, +WHICH = 77,371,252,455,336,267,181,195,264; SEARCHED THROUGH 2^30,739,014. + +ITEM (SCHROEPPEL): +TAKE AS MANY NUMBERS AS POSSIBLE FROM 1 TO N SUCH THAT NO 3 ARE +IN ARITHMETIC PROGRESSION. CONJECTURE: THE DENSITY OF SUCH SETS APPROACHES ZERO +AS N APPROACHES INFINITY; IN FACT, IT GOES AS N^((LN 2)/(LN 3)). +XX.XX IS A KNOWN SOLUTION FOR N = 5 +XX.XX....XX.XX IS A KNOWN SOLUTION FOR N = 14 +CONJECTURE THAT XX.XX JUST KEEPS GETTING COPIED. +IF THE N^((LN 2)/(LN 3)) CAN BE PROVED, IT FOLLOWS THAT THERE ARE INFINITELY MANY +PRIMES P1, P2, P3 IN ARITHMETIC PROGRESSION, SINCE PRIMES ARE MUCH MORE +COMMON THAN N^((LN 2)/(LN 3)). + +ITEM (SCHROEPPEL): +PROBLEM: HOW MANY SQUARES HAVE NO ZEROS IN THEIR DECIMAL EXPRESSION? + +ITEM (GOSPER): +THE FOLLOWING WAS CALCULATED WITH PETER SAMSON'S +STRING HANDLING INTERPRETIVE TRANSLATOR PROGRAM, "STRING." +THE PROBABILITY OF AT LEAST ONE OF THE DECIMAL DIGITS NOT APPEARING +IN A RANDOM STRING OF DECIMAL DIGITS OF LENGTH N IS: +N PROBABILITY +0 THROUGH 9 1/1 +10 1,561,933/1,562,500 +11 3,118,763/3,125,000 +12 31,056,653/31,250,000 +13 61,608,109/62,500,000 +14 949,886,851/976,562,500 +15 2,981,405,549/3,125,000,000 +16 145,264,092,067/156,250,000,000 +17 281,220,490,739/312,500,000,000 +18 135,207,403,123,699/156,250,000,000,000 +19 258,374,516,368,781/312,500,000,000,000 +20 2,453,945,865,008,123/3,125,000,000,000,000 +21 4,635,475,839,633,883/6,250,000,000,000,000 +22 544,486,732,876,350,041/781,250,000,000,000,000 +23 509,277,073,221,163,409/781,250,000,000,000,000 +24 189,739,871,255,887,679/312,500,000,000,000,000 +25 70,424,508,530,924,423/125,000,000,000,000,000 +26 1,628,170,420,719,167,436,233/3,125,000,000,000,000,000,000 +27 3,002,290,440,846,272,623,291/6,250,000,000,000,000,000,000 +28 27,606,049,370,460,038,309,269/62,500,000,000,000,000,000,000 +50 APPROXIMATELY 0.0508976470723677636754403338329777 + +ITEM (GOSPER): +THE VARIANCE OF A PSEUDO-GAUSSIAN DISTRIBUTED RANDOM VARIABLE MADE BY ADDING +T INDEPENDENT, UNIFORMLY DISTRIBUTED RANDOM INTEGER VARIABLES +WHICH RANGE FROM 0 TO N-1, INCLUSIVE, IS T((N^2 - 1)/12). + +ITEM (SCHROEPPEL): +THE JOYS OF 239 ARE AS FOLLOWS: +PI = 16 ARCTAN (1/5) - 4 ARCTAN (1/239), +WHICH IS RELATED TO THE FACT THAT 2 X 13^4 - 1 = 239^2. +239 NEEDS 4 SQUARES (THE MAXIMUM) TO EXPRESS IT. +239 NEEDS 9 CUBES (THE MAXIMUM, SHARED ONLY WITH 23) TO EXPRESS IT. +239 NEEDS 19 FOURTH POWERS (THE MAXIMUM) TO EXPRESS IT. +(ALTHOUGH 239 DOESN'T NEED THE MAXIMUM NUMBER OF FIFTH POWERS.) +1/239 = .00418410041841..., WHICH IS RELATED TO THE FACT THAT +1,111,111 = 239 X 4,649. +THE 239TH MERSENNE NUMBER, 2^239 - 1, IS KNOWN COMPOSITE, BUT NO FACTORS ARE KNOWN. +239 = 11101111 BASE 2. +239 = 22212 BASE 3. +239 = 3233 BASE 4. +AND 239 IS PRIME, OF COURSE. + +ITEM (GOSPER, SALAMIN): +PI = 16 ARCTAN (1/5) - 4 ARCTAN (1/239) +PI = 28 ARCTAN (3/79) + 20 ARCTAN (29/278) +PI = 48 ARCTAN (3/79) + 20 ARCTAN (1457/22049) +SEE ALSO RAMANUJAN'S COLLECTED PAPERS FOR A SERIES FOR +1/PI (6 DECIMAL PLACES PER TERM) AND A SERIES FOR +1/(PI SQRT 8) (8 DECIMAL PLACES PER TERM). + +ITEM (BEELER): +THE "LENGTH" OF AN N-DIGIT DECIMAL NUMBER IS DEFINED AS THE NUMBER +OF TIMES ONE MUST ITERATIVELY FORM THE PRODUCT OF ITS DIGITS +UNTIL ONE OBTAINS A ONE-DIGIT PRODUCT (SEE TECHNOLOGY REVIEW PUZZLE CORNER, +DECEMBER 1969 AND APRIL 1970). FOR VARIOUS N, THE FOLLOWING SHOWS +THE MAXIMUM "LENGTH", AS WELL AS HOW MANY DISTINCT NUMBERS +(PERMUTATION GROUPS OF N DIGITS) THERE ARE: +N MAX L DISTINCT + 2 4 54 + 3 5 219 + 4 6 714 + 5 7 2,001 + 6 7 5,004 + 7 8 11,439 + 8 9 24,309 + 9 9 48,619 +10 10 92,377 +11 10 167,959 +12 10 293,929 +ALSO, FOR N = 10, 11 AND 12, A TENDENCY FOR THERE TO BE MANY +FEWER NUMBERS OF "LENGTH" = 7 IS NOTED. OTHER THAN THIS, +THE FREQUENCY OF NUMBERS OF ANY GIVEN N, THROUGH N = 12, +DECREASES WITH INCREASING "LENGTH". + +ITEM (BEELER): +THE FIBONACCI SERIES MODULO M HAS BEEN STUDIED. THIS SERIES HAS A CYCLE LENGTH L +AND WITHIN THIS CYCLE HAS SUB-CYCLES WHICH ARE BOUNDED BY ZERO MEMBERS. +THE LENGTH OF POWERS OF PRIMES SEEMS TO BE + L = (LENGTH OF PRIME) X (PRIME^(POWER - 1)) +THE LENGTH OF PRODUCTS OF POWERS OF PRIMES SEEMS TO BE + L = LEAST COMMON MULTIPLE OF LENGTHS OF POWERS OF PRIMES WHICH ARE FACTORS +THERE CAN BE ONLY 1, 2 OR 4 SUB-CYCLES IN THE CYCLE OF A PRIME. +PRIMES WITH 1 SUB-CYCLE SEEM TO HAVE LENGTHS + L = (PRIME - 1)/N, N COVERING ALL INTEGERS +PRIMES WITH 2 SUB-CYCLES SEEM TO HAVE LENGTHS + L = (PRIME - (-1)^M)/M, M COVERING + ALL INTEGERS EXCEPT 10 K + 5 FORM +PRIMES WITH 4 SUB-CYCLES SEEM TO ALWAYS BE OF FORM 4 K + 1, +AND SEEM TO HAVE LENGTHS + L = 2 (PRIME + 1)/R OR (PRIME - 1)/S, + R COVERING ALL INTEGERS OF FORM 10 K + 1, 3, 7 OR 9 + S COVERING ALL INTEGERS +AT SCHROEPPEL'S SUGGESTION, THE PRIMES HAVE BEEN SEPARATED MOD 40, +WHICH USUALLY DETERMINES THEIR NUMBER OF SUB-CYCLES: +PRIME MOD 40 SUB-CYCLES +1, 9 USUALLY 2, OCCASIONALLY 1 OR 4 (ABOUT EQUALLY) +3, 7, 23, 27 2 +11, 19, 31, 39 1 +13, 17, 33, 37 4 +21, 29 1 OR 4 (ABOUT EQUALLY) +2 (ONLY 2) 1 +5 (ONLY 5) 4 +ATTENTION WAS DIRECTED TO PRIMES WHICH ARE 1 OR 9 MOD 40 +BUT HAVE 1 OR 4 SUBCYCLES. 25 X^2 + 16 Y^2 SEEMS TO EXPRESS THOSE +WHICH ARE 9 MOD 40; (10 X +/- 1)^2 + 400 Y^2 SEEMS TO EXPRESS +THOSE WHICH ARE 1 MOD 40. +PROBLEM: CAN SOME OF THE "SEEMS" ABOVE BE PROVED? +ALSO, CAN A GENERAL TEST BE MADE WHICH WILL PREDICT EXACT LENGTH FOR ANY NUMBER? + +ITEM (BEELER): +IF S = THE SUM OF ALL INTEGERS WHICH EXACTLY DIVIDE N, INCLUDING 1 AND N, +THEN "PERFECT NUMBERS" ARE S = 2 N; +THE FIRST THREE NUMBERS WHICH ARE S = 3 N ARE: +120 = 2^3 X 3 X 5 = 1111000 BASE 2 +672 = 2^5 X 3 X 7 = 1010100000 BASE 2 +523,776 = 2^9 X 3 X 11 X 31 = 1111111111000000000 BASE 2 + +ITEM: +THERE IS A UNIQUE "MAGIC HEXAGON" OF SIDE 3: + 3 17 18 + 19 7 1 11 +16 2 5 6 9 + 12 4 8 14 + 10 13 15 + ********************************************* +AUTOMATA THHEORY +********************************************* + +ITEM (SCHROEPPEL): +A 2-COUNTER MACHINE, GIVEN N IN ONE OF THE COUNTERS, CANNOT GENERATE 2^N. +(SATURDAY, SEPTEMBER 26, 1970) + +ITEM (SCHROEPPEL): +WHAT EFFORT IS REQUIRED TO COMPUTE PI(X), THE NUMBER OF PRIMES < X? + ********************************************* +GAMES +********************************************* + +ITEM (SCHROEPPEL): +REGARDING "POKER COINS" GAME, WHOSE RULES ARE: +1 A PLAYER THROWS N; HE MUST THEN PUT ASIDE AT LEAST ONE AND RETHROW THE REST +2 THIS THROWING IS REPEATED UNTIL HE NO LONGER HAS ANY TO THROW +3 HIGHEST SCORE (DICE) OR MAXIMUM NUMBER OF HEADS (COINS) WINS +FOR POKER COINS, THE OPTIMAL STRATEGY, WITH N COINS THROWN, IS: +Z = NUMBER OF ZEROS (TAILS) + IF Z = 0, QUIT + IF Z = 1, THROW THE ZERO + IF 1 < Z < N, SAVE ONE ONE, THROW THE OTHER N-1 COINS + IF Z = N, SAVE A ZERO, THROW THE OTHER N-1 COINS +THE OPTIMAL STRATEGY FOR POKER DICE IS HARIER. + +ITEM (SCHROEPPEL): +PROBLEM: SOLVE BLACKOUT, A GAME AS FOLLOWS: +TWO PLAYERS ALTERNATE PLACING X'S ON A RECTANGULAR GRID. +NO TWO X'S MAY APPEAR ADJACENT ALONG A SIDE OR ACROSS THE DIAGONAL AT A CORNER. +THE LAST X WINS. +KNOWN BEHAVIOR: + FOR LINES, A PERIOD OF 34 IS ENTERED AFTER LINE IS ABOUT 80 LONG. + FOR L'S, THERE IS A TENDANCY TOWARD LONGER INITIAL SETTLING FOR LARGER L'S, + THEN IT SEEMS TO HAVE PERIOD 34. +THE "NUMBER" FOR A POSITION IS: + MAKE ALL POSSIBLE MOVES, EVALUATING POSITION AFTER EACH + "NUMBER" IS MINIMUM NUMBER NOT THUS ENCOUNTERED + THE NULL GAME = "NUMBER" OF ZERO = 2ND PLAYER WINS +THIS "NUMBER" ANALYSIS IS SIMILAR FOR MANY OTHER TAKE-AWAY GAMES, SUCH AS NIM. +FOR 2 BOARDS (NOT TOUCHING) PLAYED TOGETHER, "NUMBER" IS XOR OF THE 2 + NUMBERS FOR THE SEPARATE BOARDS. +ON AN ODD X ODD BOARD, THE 1ST PLAYER WINS. +ON A 4 X N BOARD, THE 2ND PLAYER WINS. +ON A 6 X 6 BOARD, THE 1ST PLAYER WINS BY PLAYING AT THE CENTER OF ONE QUARTER. + +ITEM: +BERLEKAMP OF BELL LABS HAS DONE THE 9 SQUARES (16 DOTS) DOTS GAME; +THE 2ND PLAYER WINS. + +ITEM (BEELER): +THERE IS ONLY ONE DISTINCT SOLUTION TO THE COMMERCIAL "INSTANT INSANITY" +COLORED-FACES CUBES PUZZLE, WHICH IS HOW IT COMES PACKED. + +ITEM (BEELER): +A WINDOW-DICE GAME IS AS FOLLOWS: +1 THE PLAYER STARTS WITH EACH OF NINE WINDOWS OPEN, SHOWING THE DIGITS 1 - 9. +2 ROLL TWO DICE. +3 COVER UP ANY DIGITS WHOSE SUM IS THE SUM ON THE DICE. +4 ITERATE THROWING AND CLOSING WINDOWS UNTIL THE EQUALITY OF SUMS IS IMPOSSIBLE. +5 YOUR SCORE (MINIMUM WINS) IS THE TOTAL OF OPEN WINDOWS (UNCOVERED DIGITS). +AN OPTIMUM STRATEGY HAS BEEN TABULATED. +USUALLY IT IS BEST TO TAKE THE LARGEST DIGITS POSSIBLE, BUT NOT ALWAYS; +IT ALSO DEPENDS CRITICALLY ON THE REMAINING NUMBERS. + +ITEM (BEELER): +SIM IS A GAME WHERE TWO PLAYERS ALTERNATELY DRAW LINES CONNECTING SIX DOTS. +THE FIRST PERSON TO FORM A TRIANGLE IN HIS COLOR LOSES. +THE SECOND PLAYER CAN ALWAYS WIN, AND WHETHER HIS FIRST MOVE CONNECTS WITH +THE FIRST PLAYER'S FIRST MOVE DOESN'T MATTER; FROM THERE ON, HOWEVER, +THE STRATEGY BRANCHES TO A RELATIVELY GRUESOME DEGREE. +PROBLEM: 6 DOTS IS MINIMUM TO ENSURE NO STALEMATE WHTH 2 PLAYERS; +HOW MANY DOTS ARE REQUIRED WITH 3 PLAYERS? + +ITEM (BEELER): +THE 4 X 4 GAME OF NIM IS A WIN FOR THE SECOND PLAYER, WHO ON +HIS FIRST MOVE CAN REPLY CENTER-SYMMETRICALLY UNLESS THE FIRST +PLAYER'S FIRST MOVE WAS B1 AND B2 (ANALYZED ON RLE PDP-1). + +ITEM (BEELER): +TRIANGULAR HI-Q (OR PEG SOLITAIRE) IS 15 PEGS IN A TRIANGLE. +ONE PEG IS REMOVED, AND THEREAFTER PEGS JUMP OTHERS, WHICH ARE REMOVED. +WITH PEGS NUMBERED 1 AT THE TOP, 2 AND 3 IN THE NEXT ROW, ETC., +REMOVE CAN END WITH ONLY THE PEG +1 1, 7 = 10, 13 +2 2, 6, 11, 14 +4 3 = 12, 4, 9, 15 +5 13 +REMOVING ONLY ONE, NO WAY EXISTS TO GET TO EITHER 1 + 11 + 15 (TIPS) +OR 4 + 6 + 13 (CENTERS OF SIDES). STARTING WITH PEG 1 REMOVED, +3,016 POSITIONS ARE ATTAINABLE (NOT TURNING BOARD); +THE SUM OF WAYS TO GET TO EACH OF THESE IS 10,306. +AN EXAMPLE IS: REMOVE PEG 1, THEN JUMP AS FOLLOWS: +6, 13, 10, 1, 2, 11, 14/13, 6, 12/13, 15, 7/4, 13, 4; LEAVING PEG 1. + ********************************************* +PROPOSED (SCHROEPPEL) COMPUTER PROGRAMS, +IN ORDER OF INCREASING RUNNING TIME +********************************************* + +PROBLEM: COUNT THE POLYOMINOS UP TO, SAY, ORDER 20. +FROM APPLIED COMBINATORIAL MATHEMATICS, PAGES 201 AND 213: +ORDER E. H. NOT ENCLOSING HOLES +1 1 1 +2 1 1 +3 2 2 +4 5 5 +5 12 12 +6 35 35 +7 108 107 +8 369 363 +9 1285 1248 +10 4655 4271 +11 17073 +12 63600 + +PROBLEM: SOLVE "MINICHESS", CHESS PLAYED ON A 5 X 5 BOARD WHERE EACH SIDE +HAS LOST A ROOK, KNIGHT, BISHOP, AND 3 PAWNS FROM ONE SIDE AND THE OPPONENTS ARE SHOVED +CLOSER TOGETHER (1 EMPTY ROW INTERVENING). + +PROBLEM: SOLVE THE TIGER PUZZLE, A SLIDING BLOCK PUZZLE MENTIONED IN +SCIENTIFIC AMERICAN FEBRUARY 1964, PAGES 122 - 130. + +PROBLEM: FIND SMALLEST SQUARED SQUARE (A SQUARE COMPOSED ENTIRELY OF SMALLER, +UNEQUAL SQUARES). SMALLEST KNOWN HAS 24 SMALL SQUARES +(MARTIN GARDNER'S SCIENTIFIC AMERICAN BOOK, VOL. 2, PAGE 206). + +PROBLEM: LIST (THAT IS, COUNT) THE SEMIGROUPS OF 7 ELEMENTS; +ALSO, THE GROUPS OF 256 ELEMENTS. + +PROBLEM: COMPUTE AS A FUNCTION OF N THE LARGEST SIZE FOR N SMALL CIRCLES +TO BE PLACED INSIDE ONE OF UNIT SIZE. FOR EXAMPLE, THE SIZE IS THE +SAME FOR N = 6 AND N = 7 (GOSPER); ARE THERE ANY OTHER SUCH +PHENOMENA? + +PROBLEM: SOLVE PENTOMINOS ON AN 8 X 8 CHECKERBOARD GAME(S). RULES: +1 THE CHECKERBOARD IS FOR AID IN ORIENTING ONLY; BLACK AND WHITE ARE THE SAME. +2 THE TWO PLAYERS MAY EACH HAVE A FULL COMPLEMENT OF 12 PENTOMINOS, OR + THEY MAY "CHOOSE UP" THEIR HALF OF ONE SET. +3 THE LAST PLAYER TO PLACE A PENTOMINO WINS. + +PROBLEM: WITH REGARD TO DISSECTION THEOREMS, THE FOLLOWING ARE KNOWN: +A TRIANGLE INTO A SQUARE, 4 PIECES (PROVEN MINIMAL) +A PENTAGON INTO A SQUARE, 6 PIECES (BEST KNOWN) +ETC. ("GEOMETRIC DISSECTIONS" BY HARRY LINDGREEN, SCIENTIFIC AMERICAN NOVEMBER 1961). +A PROGRAM CAN PROBABLY CHECK THE KNOWN DISSECTIONS FOR MINIMALITY! + +PROBLEM: FIND THE NUMBER OF DOMINO COVERINGS FOR VARIOUS OBJECTS. +FOR EXAMPLE, AN ASYMPTOTIC FORMULA IS KNOWN FOR RECTANGLES; +ALSO, ON A SQUARE BOARD, IF SIDE MOD 4 = 0, COVERINGS APPEARS TO BE A SQUARE; +ON A SQUARE BOARD, IF SIDE MOD 4 = 2, COVERINGS APPEARS TO BE TWICE A SQUARE. +PROBLEM: ANALYZE GIVEAWAY CHESS, WHICH IS AS FOLLOWS: +1 CAPTURES MUST BE MADE, ALTHOUGH YOU CAN CHOOSE WHICH CAPTURE TO MAKE +2 PAWNS MUST BE PROMOTED TO QUEENS +3 KING IS JUST ANOTHER PIECE +4 PLAYER TO GIVE AWAY ALL PIECES FIRST WINS + +PROBLEM: ANALYZE "ESCALATION CHESS", WHERE WHITE GETS 1 MOVE, BLACK 2, WHITE 3, ETC. + +PROBLEM: IN THE GAME "4 PAWNS", ONE SIDE HAS 4 PAWNS, A KING, AND +TWO MOVES TO THE OTHER'S ONE. PROVE THE PAWNS WIN. + +PROBLEM: SOLVE TIC-TAC-TOE ON A 4 X 4 X 4 BOARD. + +PROBLEM: SOLVE SCARNE'S GAME, "TECO," WHICH IS PLAYED ON A 5 X 5 BOARD BY TWO PLAYERS WHO +ALTERNATE PLACING, ONE AT A TIME, THEIR 4 COUNTERS EACH, AFTER WHICH +THE COUNTERS ARE MOVED AROUND (INCLUDING DIAGONALLY). 4 IN A ROW OR SQUARE WINS. + +PROBLEM: SOLVE CHECKERS (COMPUTING TIME CURRENTLY ESTIMATED (SCHROEPPEL) AT 1 YEAR). + +PROBLEM: SOLVE HEX ON LARGE BOARDS (11 TO 23 ON A SIDE); THROUGH ORDER 7 HAVE +BEEN ANALYZED BY HAND. THERE IS A PROOF THAT IN GAMES WHERE +HAVING AN EXTRA MOVE CAN NEVER (REPEAT: NEVER) HURT YOU, THE WORST +THE FIRST PLAYER CAN BE FORCED TO DO IS DRAW. +THUS, WITH HEX, IN WHICH THERE IS NO DRAW, THE FIRST PLAYER CAN ALWAYS WIN. + ********************************************* +CONTINUED FRACTIONS +********************************************* + +ITEM (SCHROEPPEL): +SIMPLE PROOFS THAT CERTAIN CONTINUED FRACTIONS ARE SQRT 2, SQRT 3, ETC. +PROOF FOR SQRT 2: + X = [1, 2, 2, 2, ...] + (X-1)(X+1) = [0, 2, 2, 2, ...] * [2, 2, 2, 2, ...] = 1 + X^2 - 1 = 1 + X = SQRT 2 +PROOF FOR SQRT 3: + Y = [1, 1BAR, 2BAR] + (Y + 1)(Y - 1) = [2, 1BAR, 2BAR] * [0, 1BAR, 2BAR] + = 2 * [1, 2BAR, 1BAR] * [0, 1BAR, 2BAR] = 2 + Y^2 - 1 = 2 + Y = SQRT 3 +SIMILAR PROOFS EXIST FOR SQRT 5 AND SQRT 6; BUT SQRT 7 IS HAIRY. + +ITEM (SCHROEPPEL): +THE CONTINUED FRACTION EXPANSION OF THE POSITIVE MINIMUM OF THE GAMMA FUNCTION +(ABOUT 1.46) IS [1, 2, 6, 63, 135, 1, 1, 1, 1, 4, 1, 43, ...]. + +ITEM (SCHROEPPEL): +THE VALUE OF A CONTINUED FRACTION WITH PARTIAL QUOTIENTS INCREASING IN +ARITHMETIC PROGRESSION IS +[A+D, A+2D, A+3D, ...] = (I SUB A/D OF (2/D)/(I SUB 1+(A/D) OF (2/D) +WHERE THE I'S ARE BESSEL FUNCTIONS. +A SPECIAL CASE IS [1, 2, 3, 4, ...] = (I SUB 0 (2))/(I SUB 1 (2)). + ********************************************* +GROUP THEORY +********************************************* + +ITEM (KOMOLGOROFF, MAYBE?): +GIVEN A SET OF REAL NUMBERS, HOW MANY SETS CAN YOU GET USING ONLY +CLOSURE AND COMPLEMENT? ANSWER: 14. IF MORE DIMENSIONS ARE ALLOWED, +MORE CAN BE GOTTEN. + +ITEM (SCHROEPPEL): +AS OPPOSED TO THE USUAL FORMULATION OF A GROUP, WHERE YOU ARE GIVEN +1 THERE EXISTS AN I SUCH THAT A * I = I * A = A, AND +2 FOR ALL A, B AND C, (A * B) * C = A * (B * C), AND +3 FOR EACH A THERE EXISTS AN ABAR SUCH THAT A * ABAR = ABAR * A = 1, AND +4 SOMETIMES YOU ARE GIVEN THAT I AND ABAR ARE UNIQUE +IF INSTEAD YOU ARE GIVEN A * 1 = A AND A * ABAR = 1, THEN +THE ABOVE RULES CAN BE DERIVED. BUT IF YOU ARE GIVEN +A * 1 = A AND ABAR * A = 1, THEN +SOMETHING VERY MUCH LIKE A GROUP, BUT NOT(!) A GROUP, RESULTS. +FOR EXAMPLE, EVERY ELEMENT IS DUPLICATED. + +ITEM (GOSPER): +THE HAMILTONIAN PATHS MADE ONLY OF SWAP (SWAP ANY SPECIFIC PAIR) AND ROTATE +FOR N ELEMENTS ARE AS FOLLOWS: +N PATHS + REVERSES +2 2 + 0 +3 2 + 2 +4 3 + 3, NAMELY: + SRR RSR SRR RSR RRS RSR RSR RR + RSR SRR RSR RRS RSR RRS RSR RR + SRR RSR RRS RRS RSR RRS RRR SR +PROBLEM: A QUESTIONABLE PROGRAM SAID THERE ARE NONE FOR N = 5; IS THIS SO? + +ITEM (SCHROEPPEL): +ANY PERMUTATION ON 72 BITS CAN BE CODED WITH A ROUTINE +CONTAINING ONLY THE PDP-6/10 INSTRUCTIONS "ROT" AND "ROTC". + ORDER IAMONDS OMINOS HEXA'S SOMA-LIKE + 1 1 1 1 1 + 2 1 1 1 1 + 3 1 2 3 2 + 4 3 5 7 8 + 5 4 12 22 29 + 6 12 35 + 7 24 + 8 66 + 9 160 +10 448 +POLYOMINOS OF ORDER 1, 2 AND 3 CANNOT FORM A RECTANGLE . +ORDERS 4 AND 6 CAN BE SHOWN TO FORM NO RECTANGLES BY A CHECKERBOARD COLORING. +ORDER 5 HAS SEVERAL BOARDS AND ITS SOLUTIONS ARE DOCUMENTED +(COMMUNICATIONS OF THE ACM, OCTOBER 1965): +BOARD DISTINCT SOLUTIONS +3 X 20 2 +4 X 15 368 +5 X 12 1010 +6 X 10 2339 (VERIFIED) +TWO 5 X 6 -- 2 +8 X 8 WITH 2 X 2 HOLE IN CENTER -- 65 + +AN ORDER-4 HEXAFROB SOLUTION: + + A A A A B C C + D E B B C F C + D E E B F G G + D D E F F G G +ORDER-6 IAMOND BOARDS AND SOLUTION COUNTS: + SIDE 9 TRIANGLE WITH INVERTED SIDE 3 TRIANGLE IN CENTER REMOVED -- NO SOLUTIONS +TRAPEZOID, SIDE 6, BASES 3 & 3+6 -- NO SOLUTIONS +TWO TRIANGLES OF SIDE 6 -- NO SOLUTIONS +TRAPEZOID, SIDE 4, BASES 7 & 7+4 -- 76. DISTINCT SOLUTIONS +PARALLELOGRAM, BASE 6, SIDE 6 -- 156. DISTINCT SOLUTIONS +PARALLELOGRAM, BASE 4, SIDE 9 -- 37. DISTINCT SOLUTIONS +PARALLELOGRAM, BASE 3, SIDE 12. -- NO SOLUTIONS +TRIANGLE OF SIDE 9 WITH TRIANGLES OF SIDE 1, 2 AND 2 REMOVED FROM ITS CORNERS + (A COMMERCIAL PUZZLE) -- 5885. DISTINCT SOLUTIONS +WITH SOMA-LIKE PIECES, ORDERS 1, 2 AND 3 DO NOT HAVE INTERESTING BOXES. +ORDER 4 HAS 1390 DISTINCT SOLUTIONS FOR A 2 X 4 X 4 BOX. +1124 OF THESE HAVE THE FOUR-IN-A-ROW ON AN EDGE; THE REMAINING +266 HAVE THAT PIECE INTERNAL. 320 SOLUTIONS ARE DUE TO VARIATIONS +OF TEN DISTINCT SOLUTIONS DECOMPOSABLE INTO TWO 2 X 2 X 4 BOXES. +A SOMA-LIKE 2 X 4 X 4 SOLUTION: + + AAAA BBHH + BCCC BHHC + DDDE FGGE + FDGE FFGE + EXAMPLES OF ORDER-6 IAMOND SOLUTIONS: + + DD DDD DDF FFF FLL LLI III II + D DDD DDD DFF FFL LLL LII III I + AA ABD DJJ JFF FFL LLL LKK KKI II + A AAB BDJ JJF FFF FLL LLK KKK KII I + AA ABB BCJ JJJ JJE EEK KKK KGG GHH HH + A AAB BBC CJJ JJJ JEE EKK KKG GGH HHH H + AA ABB BBC CCC CCC CEE EEE EGG GGG GHH HH +A AAB BBB BCC CCC CCE EEE EEG GGG GGH HHH H + + GG GGG GRR RRR RRG GGG G + G GGG GGR RRR RRR RGG GG + GG GYY YGG GGR RYG GGG G + G GGY YYG GGG GRY YGG GG + RR RRY YYG GGG GYY YYY Y + R RRR RYY YGG GGY YYY YY + GG GRR RYY YBB BBB BRY Y + G GGR RRY YYB BBB BBR RY + GG GYR RGG GGG GBB BRR R + G GGY YRG GGG GGB BBR RR + GG GYY YYY YYG GGR RRR R +G GGY YYY YYY YGG GRR RR + +R RRR RCC CCC CCF FFF FTT TTG GGG GG + RR RRC CCC CCC CFF FFT TTT TGG GGG G + R RRR RUU UUC CFF FFT TDT TTG GGA AA + RR RRU UUU UCF FFF FTD DTT TGG GAA A + U UUU UEE EXX XXD DDD DSS SSS SBA AA + UU UUE EEX XXX XDD DDS SSS SSB BAA A + E EEE EEX XXX XDD DSS SBB BBB BBA AA + EE EEE EXX XXD DDS SSB BBB BBB BAA A + + JF FFF FCG GGG GGA AA + J JFF FFC CGG GGG GAA A + JJ JFF FFC CCG GGH HHH HA +J JJF FFF FCC CGG GHH HHA A +J JJJ JEE EDC CCC CIH HAA A + JJ JJE EED DCC CCI IHA AA + E EEE EED DDD DII IAA A + EE EEE EDD DDI IIA AA + B BBB BKD DDI IIA A + BB BBK KDD DII IA + B BBK KKK KKI I + BB BKK KKK KI + B BLL LLK K + BL LLL LK + L LLL L + LL LL + ********************************************* +TOPOLOGY +********************************************* + +ITEM (SCHROEPPEL): +TESSELATING THE PLANE WITH POLYOMINOS: +THROUGH ALL HEXOMINOES, THE PLANE CAN BE TESSELATED WITH EACH PIECE +(WITHOUT EVEN FLIPPING ANY OVER). ALL BUT THE FOUR HEPTOMINOS BELOW CAN +TESSELATE THE PLANE, AGAIN WITHOUT BEING FLIPPED OVER. THUS, FLIPPING +DOES NOT BUY YOU ANYTHING THROUGH ORDER 7. (THERE ARE 108 HEPTOMINOS). + H H HHH H H + HHHHH H H HHHH HHHH + HH H H + H H + +ITEM: +ALTHOUGH NOT NEW, THE FOLLOWING COLORING NUMBER (CHROMATIC NUMBER) +MAY BE USEFUL TO HAVE AROUND: + N = [[(7 + SQRT (1 + 48 * HOLES))/2]] +WHERE N IS THE NUMBER OF COLORS REQUIRED TO COLOR ANY MAP ON AN OBJECT WHICH HAS +"HOLES" NUMBER OF HOLES (NOTE: PROOF NOT VALID FOR HOLES = 0). +FOR EXAMPLE: +A DONUT (HOLES = 1) REQUIRES 7 COLORS TO COLOR MAPS ON IT. +A 17-HOLE FROB REQUIRES 17 COLORS. +AN 18-HOLE FROB REQUIRES 18 COLORS. + +ITEM (GOSPER): +THE FOLLOWING TRANSFORMATION MAPS A BIT STREAM INTO A POINT (X,Y) +ON THE INFINITE LIMIT OF A SPACE FILLING CURVE: + ********************************************* +SERIES +********************************************* + +ITEM (SCHROEPPEL & GOSPER): +THE SUM FROM N = 0 OF [N!N!/(2N)!] = 4/3 + 2*PI/9*SQRT 3. +PROBLEM: EVALUATE IN CLOSED FORM THE SUM FROM N = 0 +OF [N!N!N!/(3N)!]. THIS SUM HAS BEEN SHOWN EQUAL TO THE FOLLOWING: +THE INTEGRAL FROM 0 TO 1 OF (P + Q ARCCOS (R))DT +WHERE + 2 (8 + 7 T^2 - 7 T^3) +P = --------------------- + (4 - T^2 + T^3)^2 +AND + 4 (T - T^2) (5 + T^2 - T^3) +Q = ------------------------------------------------ + (4 - T^2 + T^3)^2 SQRT ((4 - T^2 + T^3) (1 - T)) +AND + T^2 - T^3 +R = 1 - --------- + 2 + + +ITEM (HENRY COHEN): +GAMMA = - LN X + X - X^2/2*2! + X^3/3*3! - X^4/4*4! ... + ERROR +WHERE ERROR IS OF THE ORDER OF (E^-X)/X. + +ITEM (SCHROEPPEL): +EULER'S SERIES ACCELERATION METHOD APPLIED TO PI/4 GIVES +PI/4 = SUM [2^(N-1)(N!)^2/(2N+1)!] + +ITEM (SCHROEPPEL): +PROBLEM: SQUARE SOME SERIES FOR PI TO GIVE THE SERIES +PI/6 = 1/1^2 + 1/2^2 + ... = SUM [1/N^2]. + +ITEM (SCHROEPPEL): +CONSIDER SUM [1/N^2] => + SUM [1/(N-1/2)-1/(N+1/2)] + SUM [1/N^2-1/(N^2-1/4)] + = 2 - SUM [1/((4N^2-1)*N^2)] +TAKE THE LAST SUM AND RE-APPLY THIS TRANSFORMATION. +THIS MAY BE A WINNER FOR COMPUTING THE ORIGINAL SUM. +FOR EXAMPLE, THE NEXT ITERATION GIVES + 31/18 - SUM [9/(N^2)(4N^2-1)(25N^4+5N^2+9)] + WHERE THE DENOMINATOR ALSO = (N^2)(2N+1)(2N-1)(5N^2+5N+3)(5N^2-5N+3) + +ITEM (SCHROEPPEL): +THE "PARITY NUMBER" IS THE SUM (0 TO INFINITY) OF (2^-N)*(PARITY OF N) +WHERE THE PARITY OF N IS 0 OR 1. THE PARITY NUMBER'S VALUE IS BETWEEN +0 AND 1. IT CAN BE WRITTEN IN STAGES BY TAKING THE PREVIOUS STAGE, COMPLEMENTING, +AND APPENDING TO THE PREVIOUS STAGE: + 0. + 0.1 + 0.110 + 0.1101001 + 0.110100110010110 + 0.1101001100101101001... RADIX 2 +OR, FASTER (GOSPER), BY SUBSTITUTING IN THE STRING AT ANY STAGE: + THE STRING ITSELF FOR ZEROS, AND + THE COMPLEMENT OF THE STRING FOR ONES. +FOR HEXADECIMAL FREAKS, HALF THE PARITY NUMBER IS .6996966996696996... +IT IS CLAIMED (PERHAPS PROVEN BY THUE?) THAT THE PARITY NUMBER IS TRANSCENDENTAL. +IN ADDITION, + 2*(PARITY NUMBER) = 2-(1/2)*(3/4)*(15/16)*(255/256)*(65535/65536)*... + ********************************************* +FLOWS AND ITERATED FUNCTIONS +********************************************* + +ITEM (SCHROEPPEL): +AN ANALYTIC FLOW FOR NEWTON'S METHOD: +DEFINE F(X) BY (X^2+K)/2X; THEN + (X+SQRT K)^2^N + (X-SQRT K)^2^N +F(F(F(...(X)))) = SQRT K ------------------------------- + [N TIMES] (X+SQRT K)^2^N - (X-SQRT K)^2^N + +ITEM (SCHROEPPEL): +SUPPOSE Y SATISFIES A DIFFERENTIAL EQUATION OF THE FORM + P(X)Y(NTH DERIVATIVE) + ..... + Q(X) = R(X) +WHERE P, ..... Q, AND R ARE POLYNOMIALS IN X +(FOR EXAMPLE, BESSEL'S EQUATION, X^2Y''+XY'+(X^2-N^2)Y = 0) +AND A IS AN ALGEBRAIC NUMBER +THEN Y(A) CAN BE EVALUATED TO N PLACES IN TIME PROPORTIONAL TO N(LN N)^C +WHERE IT SEEMS C ALWAYS = 1 PROBABLY. +FURTHER, E^X AND LN X CAN BE EVALUATED TO N PLACES IN SIMILAR TIME +FOR X A REAL NUMBER. +IF F(X) CAN BE EVALUATED IN SUCH TIME, SO CAN THE INVERSE OF F(X) +(BY NEWTON'S METHOD), AND THE FIRST DERIVATIVE OF F(X). + +ITEM (SCHROEPPEL): +P AND Q ARE POLYNOMIALS IN X; WHEN DOES P(Q(X)) = Q(P(X)) ? +(THAT IS, P COMPOSED WITH Q = Q COMPOSED WITH P.) +KNOWN SOLUTIONS ARE: +1 VARIOUS LINEAR THINGS +2 X TO DIFFERENT POWERS, SOMETIMES MULTIPLIED BY ROOTS OF 1 +3 P AND Q ARE EACH ANOTHER POLYNOMIAL R COMPOSED WITH ITSELF DIFFERENT NUMBERS OF TIMES +4 SOLUTIONS ARISING OUT OF THE FLOW OF X^2-2, AS FOLLOWS: + SUPPOSE X = Y+1/Y + THEN Y^N+1/Y^N CAN BE WRITTEN AS A POLYNOMIAL IN X + FOR EXAMPLE, P = THE EXPRESSION FOR SQUARES = X^2-2 (N = 2) + AND Q = THE EXPRESSION FOR CUBES = X^3-3X (N = 3) +5 REPLACE X BY Y-A, THEN ADD A TO THE ORIGINAL CONSTANTS + IN BOTH P AND Q. FOR EXAMPLE, P = X^2 AND Q = X^3 + THEN P = 1+(Y-1)^2 = Y^2-2Y+2 AND Q = 1+(Y-1)^3 + THEN P(Q) = 1+(Y-1)^6 = Q(P) + SIMILARLY, REPLACING X WITH AY+B WORKS. +6 THERE ARE NO MORE THROUGH DEGREES 3 AND 4 (CHECKED WITH MATHLAB); + BUT ARE THERE ANY MORE AT ALL? + +ITEM (SCHROEPPEL): +PROBLEM: GIVEN F(X) AS A POWER SERIES IN X WITH CONSTANT TERM = 0, +WRITE THE FLOW POWER SERIES. +FLOW SUB ZERO = X +FLOW SUB ONE = F(X) +FLOW SUB TWO = F(F(X)) +ETC. + +ITEM (SCHROEPPEL, GOSPER, HENNEMAN & BANKS): +THE "3N+1 PROBLEM" IS ITERATIVELY REPLACING N BY N/2 IF N IS EVEN +OR BY 3N+1 IF N IS ODD. KNOWN LOOPS FOR N TO FALL INTO ARE: +1 THE ZERO LOOP, 0 => 0 +2 A POSITIVE LOOP, 4 => 2 => 1 => 4 +3 THREE NEGATIVE LOOPS (EQUIVALENT TO THE 3N-1 PROBLEM WITH POSITIVE N) + -2 => -1 => -2 + -5 => -7 => -10 => -5 + -17 => -25 => -37 => -55 => -82 => -41 => -61 => -91 => -136 => -68 => -34 => -17 +IN THE RANGE -100 * 10^6 < N < 60 * 10^6, ALL N FALL INTO THE ABOVE LOOPS. +ARE THERE ANY OTHER LOOPS? DOES N EVER DIVERGE TO INFINITY? + +ITEM (SCHROEPPEL): +TAKING ANY TWO NUMBERS A AND B, FINDING THEIR ARITHMETIC MEAN AND THEIR +GEOMETRIC MEAN, AND USING THESE MEANS AS A NEW A AND B, THIS PROCESS, WHEN REPEATED, +WILL APPROACH A LIMIT WHICH CAN BE EXPRESSED IN TERMS OF ELLIPTIC FUNCTIONS. + +ITEM (SCHROEPPEL): +PROBLEM: ALTHOUGH THE REASON FOR THE CIRCLE ALGORITHM'S STABILITY IS UNCLEAR, +WHAT IS THE NUMBER OF DISTINCT SETS OF RADII? (NOTE: ALGORITHM IS +INVERTIBLE, SO ALL POINTS HAVE PREDECESSORS.) +CIRCLE ALGORITHM: + NEW X = OLD X - EPSILON * OLD Y + NEW Y = OLD Y + EPSILON * NEW(!) X + + ********************************************* +SET THEORY +********************************************* + +ITEM (GOSPER): +ALEPH-TWO FUNCTIONS SATISFY F(F(X)) = SIN X. + *********************************************PROGRAMMING HACKS +********************************************* + +ITEM (GOSPER): +TO SWAP THE CONTENTS OF TWO LOCATIONS IN MEMORY: + EXCH A,LOC1 + EXCH A,LOC2 + EXCH A,LOC1 +NOTE: LOC1 MUST NOT EQUAL LOC2! + +ITEM (GOSPER): +TO SWAP TWO BITS IN AN ACCUMULATOR: + TRCE A,BIT1 + TRCE A,BIT2 + TRCE A,BIT1 +NOTE: LAST TRCE NEVER SKIPS, AND COULD BE A TRC, BUT TRCE IS LESS FORGETABLE. +ALSO, USE TLCE OR TDCE IF THE BITS ARE NOT IN THE RIGHT HALF. + +ITEM (GOSPER): +PROVING THAT SIMPLE PROGRAMS ARE NEITHER TRIVIAL NOR +EXHAUSTED YET, THERE IS THE FOLLOWING: + TLCA 1,1(1) + ROT 1,11 + JRST .-1 +THIS MAKES PRETTY PATERNS (THAT IS, IT IS A DISPLAY HACK) +WITH THE LOW 9 BITS = Y AND THE NEXT HIGHER = X; ALSO, IT MAKES INTERESTING, +RELATED NOISES WITH A STEREO AMP HOOKED TO THE X AND Y SIGNALS. +RECCOMMENDED VARIATIONS INCLUDE: + CHANGE GOOD <1> + NONE 377767,,377767; 757777,,757757; MANY OTHERS + TLC 1,2(1) 373777,,0; 300000,,0 + TLC 1,3(1) -2,,-2; -5,,-1; -6,,-1 + ROT 1,1 7,,7; A0000B,,A0000B + ROTC 1,11 + AOJA 1,.-2 + +ITEM (SUSSMAN): +TO EXCHANGE TWO VARIABLES IN LISP WITHOUT USING A THIRD VARIABLE: + (SETQ X (PROG2 0 Y (SETQ Y X))) + +ITEM (GOSPER): +A MINIATURE SINE AND COSINE ROUTINE FOLLOWS. +COS: FADR A,[1.57079632689] +SIN: MOVM B,A + CAMG B,[.0002] + POPJ P, + FDVR A,[-3.0] + PUSHJ P,SIN + FMPR B,B + FSC B,2 + FADR B,[-3.0] + FMPRB A,B + POPJ P, + *********************************************PROGRAMMING ALGORITHMS, HEURISTICS +********************************************* + +ITEM (GOSPER): +THE "BANANA PHENOMENON" WAS ENCOUNTERED WHEN PROCESSING A TEXT OF NEWS +BY TAKING THE LAST 3 LETTERS OUTPUT, SEARCHING FOR A RANDOM +OCCURRENCE IN THE TEXT OF THAT SEQUENCE, TAKING THE LETTER FOLLOWING +THAT OCCURRENCE, TYPING IT OUT, AND ITERATING. +THIS ENSURES THAT EVERY 4-LETTER STRING OUTPUT OCCURS IN THE ORIGINAL. +THE PROGRAM TYPED BANANANANANANANA.... +WE NOTE AN AMBIGUITY IN THE PHRASE, "THE NTH OCCURRENCE OF." +IN ONE SENSE, THERE ARE FIVE 00'S IN 0000000000; IN ANOTHER, THERE ARE NINE. +THE EDITING PROGRAM TECO MAKES AN ARBITRARY DECISION OF WHICH YOU MEAN +(IT FINDS FIVE). SUCH PROGRAMS AND ROUTINES, HOWEVER, MIGHT +WELL BENEFIT FROM AN OPTION ALLOWING ONE TO SPECIFY WHICH IS MEANT. +  ð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒð0`Áƒ): +P AND Q ARE POLYNOMIALS IN X; WHEN DOES P(Q(X)) = Q(P(X)) ? +(THAT IS, P COMPOSED WITH Q = Q COMPOSED WITH P.) +KNOWN SOLUTIONS ARE: +1 VARIOUS LINEAR THINGS +2 X TO DIFFERENT POWERS, SOMETIMES MULTIPLIED BY ROOTS OF 1 +3 P AND Q ARE EACH ANOTHER POLYNOMIAL R COMPOSED WITH ITSELF DIFFERENT NUMBERS OF TIMES +4 SOLUTIONS ARISING OUT OF THE FLOW OF X^2-2, AS FOLLOWS: + SUPPOSE X = Y+1/Y + THEN Y^N+1/Y^N CAN BE WRITTEN AS A POLYNOMIAL IN X + FOR EXAMPLE, P = THE EXPRESSION FOR SQUARES = X^2-2 (N = 2) + AND Q = THE EXPRESSION FOR CUBES = X^3-3X (N = 3) +5 REPLACE X BY Y-A, THEN ADD A TO THE ORIGINAL CONSTANTS + IN BOTH P AND Q. FOR EXAMPLE, P = X^2 AND Q = X^3 + THEN P = 1+(Y-1)^2 = Y^2-2Y+2 AND Q = 1+(Y-1)^3 + THEN P(Q) = 1+(Y-1)^6 = Q(P) + SIMILARLY, REPLACING X WITH AY+B WORKS. +6 THERE ARE NO MORE THROUGH DEGREES 3 AND 4 (CHECKED WITH MATHLAB); + BUT ARE THERE ANY MORE AT ALL? + +ITEM (SCHROEPPEL): +PROBLEM: GIVEN F(X) AS A POWER SERIES IN X WITH CONSTANT TERM = 0, +WRITE THE FLOW POWER SERIES. +FLOW SUB ZERO = X +FLOW SUB ONE = F(X) +FLOW SUB TWO = F(F(X)) +ETC. + +ITEM (SCHROEPPEL, GOSPER, HENNEMAN & BANKS): +THE "3N+1 PROBLEM" IS ITERATIVELY REPLACING N BY N/2 IF N IS EVEN +OR BY 3N+1 IF N IS ODD. KNOWN LOOPS FOR N TO FALL INTO ARE: +1 THE ZERO LOOP \ No newline at end of file